Sample Timed Test (Quantitative)

1) If X / Y = 9/7, then find (Y – X) / X?

2/9

- 2/9

- 2/7

2/7

1/7

ANS: (B) (X/Y = 9/7 or Y/X = 7/9 or (Y/X) – 1 = (7/9) -1 or [(Y – X) / X] = - 2/9.)

Explanation

ANS: (B) (X/Y = 9/7 or Y/X = 7/9 or (Y/X) – 1 = (7/9) -1 or [(Y – X) / X] = - 2/9.)

2) If f(x) = x⁴ – x³ + x² – x +1, solve [f(-2) – f(2)] / 2 ?

10

-10

1

-1

21

ANS: (A) (f(-2) is when x =-2, so f(-2) = (-2)^4 – (-2)^3 + (-2)^2 – (-2) +1 = 16 – (-8) + 4 + 2 +1 = 31
Similarly, f(2) = 16 – 8 + 4 – 2 + 1 = 11
So [f(-2) – f(2)] / 2 = (31 – 11) / 2 = 10 )

Explanation

ANS: (A) (f(-2) is when x =-2, so f(-2) = (-2)^4 – (-2)^3 + (-2)^2 – (-2) +1 = 16 – (-8) + 4 + 2 +1 = 31
Similarly, f(2) = 16 – 8 + 4 – 2 + 1 = 11
So [f(-2) – f(2)] / 2 = (31 – 11) / 2 = 10 )

3) What will be W : Z?(1) W : X : Y = 2 : 9/2 : 5 and X : Y : Z = 18 : 20 : 7(2) W : X = 4 : 9, X : Y = 27 : 30, Y : Z = 60 : 21

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

ANS: (D) (As per (1), W : X : Y = 2 : 9/2 : 5 or 8 : 18 : 20 (when multiplied by 4). Also X : Y : Z = 18 : 20 : 7, so W : X : Y : Z = 8 : 18 : 20 : 7, so W : Z = 8 : 7, hence it can be solved using (a). As per (b), W : X = 4 : 9 or 8 : 18 (multiply by 2). Similarly, X : Y = 27 : 30 or 18 : 20 (multiply by 2/3). Also Y : Z = 60 : 21 or 20 : 7 (divide by 3). From This, W : X : Y : Z = 8 : 18 : 20 : 7, hence W : Z = 8 : 7. Hence it can be solved using (2) also. Hence it can be solved using both independently.)

Explanation

ANS: (D) (As per (1), W : X : Y = 2 : 9/2 : 5 or 8 : 18 : 20 (when multiplied by 4). Also X : Y : Z = 18 : 20 : 7, so W : X : Y : Z = 8 : 18 : 20 : 7, so W : Z = 8 : 7, hence it can be solved using (a). As per (b), W : X = 4 : 9 or 8 : 18 (multiply by 2). Similarly, X : Y = 27 : 30 or 18 : 20 (multiply by 2/3). Also Y : Z = 60 : 21 or 20 : 7 (divide by 3). From This, W : X : Y : Z = 8 : 18 : 20 : 7, hence W : Z = 8 : 7. Hence it can be solved using (2) also. Hence it can be solved using both independently.)

4) Which of the following is correct?

(4/7) < [(4+3)/(7-3)] < (3/4)

(4/7) < [(4-3) / (7-4)] < (3/4)

(4/7) < [(4x3) / (7x4)] < (3/4)

(4/7) < [(4+3) / (7+4)] < (3/4)

(4/7) < [(4/3) / (7/4)] < (3/4)

(Ans: (d) . (a) can be written in decimal form as 0.56

Explanation

(Ans: (d) . (a) can be written in decimal form as 0.56

5) Out of 150 students in a class, 80 students are taking at least one of the courses out of Strategy and Marketing. What is the number of students taking Strategy but not Marketing? 1. Number of students taking Strategy but not Marketing is same as number of students taking Marketing but not Strategy. 2. 36 students are taking both the courses.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient.

(Ans: (c) Number of students taking at least one of the courses (A) = number of students taking course no.1 but not the other (B) + number of students taking course no.2 but not the other (C) + number of students taking both the courses (D)
As per question, A = 80
Now taking (1), B = C, but D is not known, so (1) is not sufficient.
Taking (2), D = 36, but (2) is also not sufficient.
Taking both (1) & (2) together, 80 = 2 B + 36 (B = C)
So B = C = 22, hence sufficient, so (c) is correct.)

Explanation

(Ans: (c) Number of students taking at least one of the courses (A) = number of students taking course no.1 but not the other (B) + number of students taking course no.2 but not the other (C) + number of students taking both the courses (D)
As per question, A = 80
Now taking (1), B = C, but D is not known, so (1) is not sufficient.
Taking (2), D = 36, but (2) is also not sufficient.
Taking both (1) & (2) together, 80 = 2 B + 36 (B = C)
So B = C = 22, hence sufficient, so (c) is correct.)

6) When an item is sold at its original price, then some quantity of that item can be purchased in $100. But when the price of that item is increased by 25%, then 5 pieces less than the earlier quantity can be purchased in the same amount. Find the original price of the item?

$5

$4.5

$4

$5.5

$6

ANS: (C) (Let A be the original price and N be the initial quantity. As per question, AN = 100. In the second case, the price becomes 1.25A but the quantity becomes N-5, so 1.25A (N-5) = 100 or 1.25AN – 6.25 A =100. Since AN = 100, so 1.25*100 – 6.25 A = 100. On solving, we get A = 4, which is the original price.)

Explanation

ANS: (C) (Let A be the original price and N be the initial quantity. As per question, AN = 100. In the second case, the price becomes 1.25A but the quantity becomes N-5, so 1.25A (N-5) = 100 or 1.25AN – 6.25 A =100. Since AN = 100, so 1.25*100 – 6.25 A = 100. On solving, we get A = 4, which is the original price.)

7) If ( */32) x (*/72) = 1, find the value of * ?

24

2304

1152

48

52

ANS: (D) ( * = SQRT (32 x 72) = SQRT(2304) =48)

Explanation

ANS: (D) ( * = SQRT (32 x 72) = SQRT(2304) =48)

8) If 3% of (m + n) is 15, where m & n are positive integers and m is a double digit number. What is the least possible value of n?

490

400

489

389

401

ANS: (E) ((3/100) ( m + n) = 15, so m + n = 500. For n to be least, m must be greatest two digit number which is 99. So n = 500 – 99 = 401.)

Explanation

ANS: (E) ((3/100) ( m + n) = 15, so m + n = 500. For n to be least, m must be greatest two digit number which is 99. So n = 500 – 99 = 401.)

9) If X > Y & X < 1 & also XY = 1, then which of the following must be true?

X > -1 & Y < -1

X < -1 & Y > -1

X > 1 & Y < 0

X > 0 & Y < -1

X < 0 & Y > 0

ANS: (A) (For XY = 1, both X & Y must be negative because X Y, so if both are positive (then both will be less than 1) then XY will be For (A), if X > -1 & Y Y. For example, if X = -1/2 and Y = -2, then XY = 1 and X > Y. So (A) may be true.
For (B), X is not > Y, so (B) can not be true.
For (C), XY will be (D) is same as (C).
Similarly, for (E), XY

Explanation

ANS: (A) (For XY = 1, both X & Y must be negative because X Y, so if both are positive (then both will be less than 1) then XY will be For (A), if X > -1 & Y Y. For example, if X = -1/2 and Y = -2, then XY = 1 and X > Y. So (A) may be true.
For (B), X is not > Y, so (B) can not be true.
For (C), XY will be (D) is same as (C).
Similarly, for (E), XY

10) In a class, the boys and girls are in a ratio of 7 : 4. If two boys leave the class and seven new girls join the class, the new ratio becomes 11 : 9. Find the number of girls in the class now?

27

37

47

24

34

ANS: (A) (Let the original number of boys be B and original no. of girls be G.
Then B/G = 7/4 or 4B = 7G
Also (B – 2)/(G + 7) = 11/9 or 9B = 11G + 95
Or (7/4) x 9G = 11G + 95 (On substituting 4B = 7G)
Or G = 20. So 20 + 7 = 27 would be the number of girls now.)

Explanation

ANS: (A) (Let the original number of boys be B and original no. of girls be G.
Then B/G = 7/4 or 4B = 7G
Also (B – 2)/(G + 7) = 11/9 or 9B = 11G + 95
Or (7/4) x 9G = 11G + 95 (On substituting 4B = 7G)
Or G = 20. So 20 + 7 = 27 would be the number of girls now.)

11) If (Xⁿ⁺¹ + Yⁿ⁺¹) / (Xⁿ + Yⁿ), is the mean of X and Y, then what will be the value of n?

1

2

3

0

-1

ANS: (d) (If n = 0, then (Xn+1 + Yn+1) / (Xn + Yn) = (X1 + Y1) / (X0 + Y0)
Or (X + Y) / 2, which is the mean of X and Y.)

Explanation

ANS: (d) (If n = 0, then (Xn+1 + Yn+1) / (Xn + Yn) = (X1 + Y1) / (X0 + Y0)
Or (X + Y) / 2, which is the mean of X and Y.)

12)

What is the area of the circle in the figure?

Arc AB of length 5 cm subtends an angle of 45 degree at the centre of the circle.
Chord MN of length 9 cm subtends an angle of 90 degree at the centre of the circle.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

(Ans: (d) Area of the circle = Pi (r^2), where r is the radius of circle.
Take (1), Length of arc = 2 Pi r (angle subtended by the arc) / 360
Or 5 = 2 Pi r (45)/360 or Pi r = 20 or r = 20 x 7 / 22
With this value of radius r, we can find the area of the circle, so (1) is sufficient.
Now take (2), since chord subtends a right angle, so using Pythagoras Theorem
We get, 9^2 = r^2 + r^2 or r = 9/square root (2)
With this value of radius r, we can find the area of the circle, so (2) is also sufficient
Hence both (1) & (2) are independently sufficient, hence (d) is the answer.)

Explanation

(Ans: (d) Area of the circle = Pi (r^2), where r is the radius of circle.
Take (1), Length of arc = 2 Pi r (angle subtended by the arc) / 360
Or 5 = 2 Pi r (45)/360 or Pi r = 20 or r = 20 x 7 / 22
With this value of radius r, we can find the area of the circle, so (1) is sufficient.
Now take (2), since chord subtends a right angle, so using Pythagoras Theorem
We get, 9^2 = r^2 + r^2 or r = 9/square root (2)
With this value of radius r, we can find the area of the circle, so (2) is also sufficient
Hence both (1) & (2) are independently sufficient, hence (d) is the answer.)

13)

In the diagram, AB passes through O which is the centre of the circle. Which of the following must be true? I. Angles APB and AMB are equal and are right angles II. Sum of sides AP & PB > Twice the radius of the circle III. Sum of sides AM & MB > Twice the radius of the circle

I Only

II & III Only

III Only

II Only

I, II & III

ANS. (E) (All angles with opposite side as diameter of a circle are right angles. No side of a triangle can be more than the sum of rest two)

Explanation

ANS. (E) (All angles with opposite side as diameter of a circle are right angles. No side of a triangle can be more than the sum of rest two)

14) Whether [(A^-2a) / (A^-2b)] + [(A^2b) / (A^2a) ] is positive, where a & b are integers?(1) a = b(2) a = -b

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

ANS: (D) (On simplifying, the equation can be written as 2/ (A^(2a – 2b)).
Take (1), a = b, so denominator becomes A^0 or 1. The value of equation = 2 (which is positive). So (1) is sufficient.
Now take (2), a = -b, so equation becomes, 2/ A^(-4b) or 2 A^4b which will always be positive whether A is +ve or –ve (b is an integer). Hence (2) is also sufficient.)

Explanation

ANS: (D) (On simplifying, the equation can be written as 2/ (A^(2a – 2b)).
Take (1), a = b, so denominator becomes A^0 or 1. The value of equation = 2 (which is positive). So (1) is sufficient.
Now take (2), a = -b, so equation becomes, 2/ A^(-4b) or 2 A^4b which will always be positive whether A is +ve or –ve (b is an integer). Hence (2) is also sufficient.)

15) x is a prime number. Find x? 1. 800 < x² < 1000 2. Sum of the digits of x is an odd number.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are NOT sufficient.

ANS: (C) (Take (1). There are two prime numbers 29 & 31, whose squares are between 800 & 1000. But with (1), we can not say whether x is 29 or 31. Hence (1) is not sufficient.
Take (2). There are several prime numbers whose sum of digits is an odd integer example 23, 29, 41 etc. So (2) is also not sufficient.
Take (1) & (2) both together, x = 29, because x2 lies between 800 & 1000 and also sum of its digits is 11 an odd number. Hence both together are sufficient.)

Explanation

ANS: (C) (Take (1). There are two prime numbers 29 & 31, whose squares are between 800 & 1000. But with (1), we can not say whether x is 29 or 31. Hence (1) is not sufficient.
Take (2). There are several prime numbers whose sum of digits is an odd integer example 23, 29, 41 etc. So (2) is also not sufficient.
Take (1) & (2) both together, x = 29, because x2 lies between 800 & 1000 and also sum of its digits is 11 an odd number. Hence both together are sufficient.)

16) Is a = 0 ?(1) (a + b)² = a² + b²(2) b = 0

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

ANS: (E) (As per (1), (a + b)^2 = a^2 + b^2 + 2ab, if (a + b)^2 = a^2 + b^2, then either a or b or both are 0. So (1) is not sufficient.
As per (2), b =0, it says nothing about a, hence it is not sufficient.
As per both (1) & (2), then either a may be zero or a may have some value. Hence both together are also not sufficient. Hence (E) is the answer.)

Explanation

ANS: (E) (As per (1), (a + b)^2 = a^2 + b^2 + 2ab, if (a + b)^2 = a^2 + b^2, then either a or b or both are 0. So (1) is not sufficient.
As per (2), b =0, it says nothing about a, hence it is not sufficient.
As per both (1) & (2), then either a may be zero or a may have some value. Hence both together are also not sufficient. Hence (E) is the answer.)

17) A can finish a work in X days and B can finish the same work in Y days. In how many less days will both A & B together finish the work than A alone can ?

XY / (X+Y)

X / (X + Y)

X^2 / (X + Y)

(X + Y) / X^2

Y^2 / (X + Y)

ANS: (C) (Both A & B together will do in one day (1/X) + (1/Y) part of work.
Or (X + Y) / XY. Number of days both will finish the work in XY / (X + Y) days.
Answer = X – XY / (X + Y) = X^2 / (X + Y)

Explanation

ANS: (C) (Both A & B together will do in one day (1/X) + (1/Y) part of work.
Or (X + Y) / XY. Number of days both will finish the work in XY / (X + Y) days.
Answer = X – XY / (X + Y) = X^2 / (X + Y)

18) On Monday, a shopkeeper sold 12(1/2) kg of flour. On Tuesday, he sold 5/6 kg more flour than on Monday. On Wednesday, he sold 1(1/3) kg of flour less than on Tuesday. He then had 1/3 kg of flour left in his shop. How much kg flour did he have in the beginning?

38(1/6)

37(1/3)

37(1/2)

38(1/3)

37(1/6)

ANS: (A) (Flour sold on Monday = 25/2 kg
Flour sold on Tuesday = 25/2 + 5/6 = 80/6 = 40/3
On Wednesday = 40/3 – 4/3 = 36/3 = 12
Total Flour = 25/2 + 40/3 + 12 + 1/3 = (75 + 80 + 72 + 2) /6
Or 229/6 = 38(1/6))

Explanation

ANS: (A) (Flour sold on Monday = 25/2 kg
Flour sold on Tuesday = 25/2 + 5/6 = 80/6 = 40/3
On Wednesday = 40/3 – 4/3 = 36/3 = 12
Total Flour = 25/2 + 40/3 + 12 + 1/3 = (75 + 80 + 72 + 2) /6
Or 229/6 = 38(1/6))

19) Find the value of {a⁶b³– [a¹²b⁹]^1/3} / { [ a²b]³ – [a⁸b¹²]^1/4}(1) a = 3(2) b = 1

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

ANS: (A) (First simplify the expression to the maximum extent possible. (Remember that (2^a)^b = 2^ab
We can rewrite that as, {(a^6)(b^3) – (a^4)(b^3) / { (a^6)(b^3) – (a^2)(b^3)}
Or {(a^4)(b^3) [(a^2) – 1] / (a^2)(b^3) [(a^4) – 1]
Or (a^2) [ (a^2) -1] / [(a^4) – 1]
Or (a^2) [ (a^2) -1] / {[(a^2) – 1][(a^2) + 1]}
Or (a^2) / [(a^2) + 1] (This is the lowest form.
From (1), a = 3, the value of expression can be found. Hence (1) is sufficient.
From (2), b = 1, there is no variable b in the reduced expression. Hence (2) is not sufficient.)

Explanation

ANS: (A) (First simplify the expression to the maximum extent possible. (Remember that (2^a)^b = 2^ab
We can rewrite that as, {(a^6)(b^3) – (a^4)(b^3) / { (a^6)(b^3) – (a^2)(b^3)}
Or {(a^4)(b^3) [(a^2) – 1] / (a^2)(b^3) [(a^4) – 1]
Or (a^2) [ (a^2) -1] / [(a^4) – 1]
Or (a^2) [ (a^2) -1] / {[(a^2) – 1][(a^2) + 1]}
Or (a^2) / [(a^2) + 1] (This is the lowest form.
From (1), a = 3, the value of expression can be found. Hence (1) is sufficient.
From (2), b = 1, there is no variable b in the reduced expression. Hence (2) is not sufficient.)

20) What is the least value of n such that 1/5ⁿ< 0.001 ?

3

4

5

6

2

ANS: (C) (1/ 5^n 1000
Now 5^4 = 625 & 5^5 = 3125 , so for n =5, 5^n > 1000.)

Explanation

ANS: (C) (1/ 5^n 1000
Now 5^4 = 625 & 5^5 = 3125 , so for n =5, 5^n > 1000.)

21) Is M an even or odd integer?(1) 4M + 3M is even.(2) 3M² + 3M³ is even.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

ANS: (A) (Here we need to understand that multiplication of even with even no. is always even, even with odd no. is also even and odd with odd no. is always odd. Also addition of even with even no. is even, even with odd is odd and odd with odd is even.
As per (1), 4M + 3M is even, so either 4M & 3M both are even or both are odd. Since 4M is even so 3M is also even. For 3M to be even, M has to be even only. Hence (1) is sufficient.
As per (2), 3M2 + 3M3 is even. We know that multiplication of even & even is even and also odd & odd is odd, so square or cube does not change the nature of an integer. Here both the terms can be either even or odd. If M is even, then the expression becomes even. When M is odd, then also the expression becomes even (both the terms are odd), hence M can be even or odd. Hence (2) is not sufficient.)

Explanation

ANS: (A) (Here we need to understand that multiplication of even with even no. is always even, even with odd no. is also even and odd with odd no. is always odd. Also addition of even with even no. is even, even with odd is odd and odd with odd is even.
As per (1), 4M + 3M is even, so either 4M & 3M both are even or both are odd. Since 4M is even so 3M is also even. For 3M to be even, M has to be even only. Hence (1) is sufficient.
As per (2), 3M2 + 3M3 is even. We know that multiplication of even & even is even and also odd & odd is odd, so square or cube does not change the nature of an integer. Here both the terms can be either even or odd. If M is even, then the expression becomes even. When M is odd, then also the expression becomes even (both the terms are odd), hence M can be even or odd. Hence (2) is not sufficient.)

22) If a is a root of the equation x² + m² – (n – x)², where x is the only variable, then find the value of a?

(n^2 – m^2) / 2n

(m^2 – n^2) / 2m

(n^2 – m^2) / 2

2n / (m^2 – n^2)

2m / (m^2 – n^2)

ANS: (A) (a is a root means that if we replace x by a, then the expression becomes equal to zero.
So a^2 + m^2 – (n – a)^2 = 0
Or a^2 + m^2 – (n^2 + a^2 – 2na) = 0
Or a^2 + m^2 – n^2 – a^2 + 2na = 0
Or a = (n^2 – m^2) / 2n )

Explanation

ANS: (A) (a is a root means that if we replace x by a, then the expression becomes equal to zero.
So a^2 + m^2 – (n – a)^2 = 0
Or a^2 + m^2 – (n^2 + a^2 – 2na) = 0
Or a^2 + m^2 – n^2 – a^2 + 2na = 0
Or a = (n^2 – m^2) / 2n )

23) Workers A, B & C can finish a work in 12, 16 and 10 days respectively. If on first day, A is assisted by B and by C on the next day alternately, on which day the work will be finished?

6

7

8

9

10

ANS: (B) (A can complete 1/12 part of work in one day, similarly B can finish 1/16 and C can finish 1/10.
On first day, work completed = 1/12 + 1/16 = 7/48 part.
On second day, work completed = 1/12 + 1/10 = 11/60 part.
In initial 2 days, total work completed = 7/48 + 11/60 = 79/240.
In 6 days, total work done = 3 x 79/240 = 237/240. So on 7th day, the work will be finished.)

Explanation

ANS: (B) (A can complete 1/12 part of work in one day, similarly B can finish 1/16 and C can finish 1/10.
On first day, work completed = 1/12 + 1/16 = 7/48 part.
On second day, work completed = 1/12 + 1/10 = 11/60 part.
In initial 2 days, total work completed = 7/48 + 11/60 = 79/240.
In 6 days, total work done = 3 x 79/240 = 237/240. So on 7th day, the work will be finished.)

24) If operation * is such that A * B = A³ + B³, is M * N < 0 ?1. (M/N) < 1 2. -2 < M < 2 and N > 2

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

(Ans: (b) Take (1), M & N can have any value +ve or –ve. If both are +ve, then value of function M * N is > 0, but if both –ve then M * N is Now take (2), if 0 0 because N is also +ve.
If M is –ve say -1.99 (almost maximum –ve possible), then M^3 = - 7.88. Also let us take N = 2.01 (almost minimum value), so N^3 = 8.12. So M^3 + N^3 is > 0 .
Even if we take M = -1.99999 & N = 2.000001, then M will be slightly more than -8 and N will be slightly more than 8.
So M3 + N3 is > 0.
So as per (2), for all possible values of M & N, M * N is > 0, hence (2) is sufficient to solve the problem. Hence (b) is the answer.)

Explanation

(Ans: (b) Take (1), M & N can have any value +ve or –ve. If both are +ve, then value of function M * N is > 0, but if both –ve then M * N is Now take (2), if 0 0 because N is also +ve.
If M is –ve say -1.99 (almost maximum –ve possible), then M^3 = - 7.88. Also let us take N = 2.01 (almost minimum value), so N^3 = 8.12. So M^3 + N^3 is > 0 .
Even if we take M = -1.99999 & N = 2.000001, then M will be slightly more than -8 and N will be slightly more than 8.
So M3 + N3 is > 0.
So as per (2), for all possible values of M & N, M * N is > 0, hence (2) is sufficient to solve the problem. Hence (b) is the answer.)

25)

The equation of AB is (1/2) y = -x + 7/2. If CD is perpendicular to AB, what may be the equation of CD?

2y – x = 6

X – 2y = 1/2

2y + 2x = 5

Y + 2x = 7

2x – y = 5

ANS: A (The above equation can be written as y = -2x + 7
The slopes of perpendicular lines are negative inverse. So the slope of CD will be 1/2.
The equation of line having slope of 1/2 will be the answer. Obviously only (A) has a slope of 1/2.)

Explanation

ANS: A (The above equation can be written as y = -2x + 7
The slopes of perpendicular lines are negative inverse. So the slope of CD will be 1/2.
The equation of line having slope of 1/2 will be the answer. Obviously only (A) has a slope of 1/2.)

26)

Area of the rectangle ABCD is 196 square cm and area of triangle BGC is 1/4^th the area of rectangle ABCD. What is the difference in areas of shaded (cross-lined) and unshaded portion in the figure?1. Area of triangle ADE is same as area of triangle CGF.2. The ratio of areas of triangle ADE to triangle BGC is same as the ratio of areas of triangle CGF to triangle BGC.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

(Ans: (e) To find the area of shaded portion and unshaded portion, we need the areas of quadrilateral ABCE and triangles ADE, BGC and CGF. As per the question, only the area of triangle BGC (1/4th of 196 = 49) and the area of rectangle are known, rest areas are not known. As per statement (1), we can not calculate the areas of ADE and CGF, hence (1) is not sufficient. As per (2), again areas of ADE and CGF can not be calculated, hence (2) is also not sufficient. If we take both (1) & (2) together, then also the areas of ADE and CGF can not be calculated. Hence (1) & (2) together are not sufficient, so answer is (e).)

Explanation

(Ans: (e) To find the area of shaded portion and unshaded portion, we need the areas of quadrilateral ABCE and triangles ADE, BGC and CGF. As per the question, only the area of triangle BGC (1/4th of 196 = 49) and the area of rectangle are known, rest areas are not known. As per statement (1), we can not calculate the areas of ADE and CGF, hence (1) is not sufficient. As per (2), again areas of ADE and CGF can not be calculated, hence (2) is also not sufficient. If we take both (1) & (2) together, then also the areas of ADE and CGF can not be calculated. Hence (1) & (2) together are not sufficient, so answer is (e).)

27) Anson has a total of $920 as currency notes in the denominations of $50, $20 and $10 with total notes being 40. What is number of notes of different denominations with Anson? 1. The ratio of notes of denominations $20 and $10 is 2 : 3. 2. The ratio of notes of denominations $50, $20 and $10 is 5 : 6 : 9.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

(Ans: (d) , First let us take statement (1),
Let $20 notes = 2X , so $10 notes = 3X and $50 notes = 40 – (2X + 3X) = 40 – 5X
Now, 50x (40 – 5X) + 20 x (2X) + 10 x (3X) = 920 (Total value of notes)
On solving, we get X = 6, so $50 notes are 10, $ 20 notes are 12 and $10 are 18. Hence (1) is sufficient.
Now take statement (2), Let $50 notes = 5X, $20 notes = 6X and $10 = 9X,
So 5X + 6X + 9X = 40, or X = 2
So $50 notes are 10, $ 20 notes are 12 and $10 are 18. Hence (2) is also sufficient.
So both (1) and (2) are independently sufficient, hence (d) is correct.)

Explanation

(Ans: (d) , First let us take statement (1),
Let $20 notes = 2X , so $10 notes = 3X and $50 notes = 40 – (2X + 3X) = 40 – 5X
Now, 50x (40 – 5X) + 20 x (2X) + 10 x (3X) = 920 (Total value of notes)
On solving, we get X = 6, so $50 notes are 10, $ 20 notes are 12 and $10 are 18. Hence (1) is sufficient.
Now take statement (2), Let $50 notes = 5X, $20 notes = 6X and $10 = 9X,
So 5X + 6X + 9X = 40, or X = 2
So $50 notes are 10, $ 20 notes are 12 and $10 are 18. Hence (2) is also sufficient.
So both (1) and (2) are independently sufficient, hence (d) is correct.)

28) A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

1/2

63/128

4/7

61/256

63/64

ANS. (a) (Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7(C)4 (Heads=4 & Tails=3) + 7(C)5 + 7(C)6 + 7(C)7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)

Explanation

ANS. (a) (Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7(C)4 (Heads=4 & Tails=3) + 7(C)5 + 7(C)6 + 7(C)7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)

29) A student wrote five different subject papers. If the marks obtained by the student in different subjects were in ratio of 5 : 6 : 7 : 8 : 9, in how many subjects did the student get more than 60% marks.(1) Each subject carried equal maximum marks.(2) Student got 70% marks overall.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

ANS: (C) (a) tells the proportion of marks of different subjects in total marks. The proportion is 5/35 or 1/7, 6/35, 7/35 or 1/5, 8/35 and 9/35. But this is not sufficient to find the answer. (b) tells the overall marks, which is 350 (if maximum marks of each is 100), but the exact proportion of each subject can not be calculated until we know the ratio of maximum marks of each subject. Hence (b) is also not sufficient. But (a) & (b) together are sufficient. The marks in 1st subject (1/7) x 350 = 50 or 50%. 2nd subject (6/35) x 350 = 60 or 60%. 3rd subject (1/5) x 350 = 70 or 70%. Other two will be 80% & 90%. So in 3 subjects, he got more than 60% marks.)

Explanation

ANS: (C) (a) tells the proportion of marks of different subjects in total marks. The proportion is 5/35 or 1/7, 6/35, 7/35 or 1/5, 8/35 and 9/35. But this is not sufficient to find the answer. (b) tells the overall marks, which is 350 (if maximum marks of each is 100), but the exact proportion of each subject can not be calculated until we know the ratio of maximum marks of each subject. Hence (b) is also not sufficient. But (a) & (b) together are sufficient. The marks in 1st subject (1/7) x 350 = 50 or 50%. 2nd subject (6/35) x 350 = 60 or 60%. 3rd subject (1/5) x 350 = 70 or 70%. Other two will be 80% & 90%. So in 3 subjects, he got more than 60% marks.)

30) A person covers 12.8 m in 5 seconds. Every second, he covers a distance which is 1/3 rd of the total distance covered till the end of the previous second. What is the distance covered after first second?

4.15 m

6.15 m

4.4 m

6.05 m

4.05 m

ANS: E (Let the person covers A m distance after first second.
So, after 1st second distance covered = A
After 2nd second distance covered = A + 1/3 of A = (4/3) A
After 3rd second distance covered = (4/3) A + (1/3) of (4/3) A = (16/9) A
After 4th second distance covered = (16/9) A + (1/3) of (16/9) A
= (64/27) A
After 5th second distance covered = (64/27) A + 1/3 of (64/27) A
= (256/81) A
S0 total distance covered in 5 seconds = (256/81)A = 12.8,
so A = 4.05m)

Explanation

ANS: E (Let the person covers A m distance after first second.
So, after 1st second distance covered = A
After 2nd second distance covered = A + 1/3 of A = (4/3) A
After 3rd second distance covered = (4/3) A + (1/3) of (4/3) A = (16/9) A
After 4th second distance covered = (16/9) A + (1/3) of (16/9) A
= (64/27) A
After 5th second distance covered = (64/27) A + 1/3 of (64/27) A
= (256/81) A
S0 total distance covered in 5 seconds = (256/81)A = 12.8,
so A = 4.05m)

31) A tank is fitted with 2 pipes, pipe A is a filling pipe and pipe B is an emptying pipe. Which pipe has higher efficiency, A or B?(1) When both pipes are open, then it takes 15 hours to fill up the tank.(2) When B is closed, then pipe A takes 5 hours to fill up the tank.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

ANS: (A) (Obviously if both the pipes are open, then if it fills up, immaterial of the time taken, then the efficiency of pipe A will be more & visa-versa.
Take (1), clearly, pipe A is more efficient.
Take (2), it only tells about the efficiency of pipe B, and nothing is mentioned about pipe A, hence (2) is not sufficient.)

Explanation

ANS: (A) (Obviously if both the pipes are open, then if it fills up, immaterial of the time taken, then the efficiency of pipe A will be more & visa-versa.
Take (1), clearly, pipe A is more efficient.
Take (2), it only tells about the efficiency of pipe B, and nothing is mentioned about pipe A, hence (2) is not sufficient.)

32)

Find the total area of the figure?1. The radius of the circle is 4 cm.2. The ratio of shaded(lined) to unshaded (unlined) parts of the triangle is 1 : 3 and ratio of unshaded (unlined) to shaded (lined) parts of the circle is 2 : 1.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

(Ans: (b) The area of the triangle = (1/2) ( base x height) = 48 sq. cm
Now take (1), The radius of the circle can give the area of the circle, but no information is available on the common area (lined). So (1) is not sufficient.
Now take (2), Area of the shaded part of the triangle = [ 1/ (1 + 3)] x 48 = 12 sq cm
And area of unshaded part of triangle = 48 – 12 = 36 sq cm
Further the area of unshaded part of circle = 2 x area of shaded part of circle ( ratio is 2 : 1)
So area of unshaded part of circle = 2 x 12 = 24 sq cm.
Hence total area of the figure = Area of triangle + area of unshaded part of circle = 48 + 24 = 72 sq cm. So (2) is sufficient. Hence (b) is correct.)

Explanation

(Ans: (b) The area of the triangle = (1/2) ( base x height) = 48 sq. cm
Now take (1), The radius of the circle can give the area of the circle, but no information is available on the common area (lined). So (1) is not sufficient.
Now take (2), Area of the shaded part of the triangle = [ 1/ (1 + 3)] x 48 = 12 sq cm
And area of unshaded part of triangle = 48 – 12 = 36 sq cm
Further the area of unshaded part of circle = 2 x area of shaded part of circle ( ratio is 2 : 1)
So area of unshaded part of circle = 2 x 12 = 24 sq cm.
Hence total area of the figure = Area of triangle + area of unshaded part of circle = 48 + 24 = 72 sq cm. So (2) is sufficient. Hence (b) is correct.)

33) Find the values of A for which A(A-7) > 8A -56 ?

7 < A < 8

A > 7 & A < 8

A > 7

A < 7 & A > 8

A > 8

ANS: (D) (A(A-7) > 8A -56 or A(A – 7) > 8 (A – 7)
Or ( A – 7) (A – 8) > 0
Multiplication of two numbers is +ve when either both are +ve ( >0) or both are –ve ( 7 & A > 8 or A Or A > 8 (when A is > 7 & 8, then higher number is taken) or A

Explanation

ANS: (D) (A(A-7) > 8A -56 or A(A – 7) > 8 (A – 7)
Or ( A – 7) (A – 8) > 0
Multiplication of two numbers is +ve when either both are +ve ( >0) or both are –ve ( 7 & A > 8 or A Or A > 8 (when A is > 7 & 8, then higher number is taken) or A

34) Is X + Y an odd number, where X & Y are integers? 1. X = Y 2. 0 < X⁵ < 25 and 50 < Y⁵ < 1000

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

(Ans: (d) , X + Y will be odd when one of X & Y is even and other is odd. If both are even or odd then X + Y will be even.
Take (1), X = Y, this means that both X & Y are same, either both even or both odd, this means that X + Y is even or is not odd. So (1) is sufficient to solve the problem.
Now take (2), 0 So X + Y = 1 + 3 = 4 , which is even not odd, so (2) is also sufficient to solve the problem. Hence the answer is (d).)

Explanation

(Ans: (d) , X + Y will be odd when one of X & Y is even and other is odd. If both are even or odd then X + Y will be even.
Take (1), X = Y, this means that both X & Y are same, either both even or both odd, this means that X + Y is even or is not odd. So (1) is sufficient to solve the problem.
Now take (2), 0 So X + Y = 1 + 3 = 4 , which is even not odd, so (2) is also sufficient to solve the problem. Hence the answer is (d).)

35) There are two circular parks adjacent to each other. The diameter of bigger park is 2.5 K times the diameter of smaller one. By how much is the area of bigger park greater than the smaller park?

5.25(K^2)

2.5 (K^2) – 1

6.25 (K^2)

6.25 (K^2) – 1

(1/4)[6.25 (K^2) – 1]

ANS: (D) (The area of the smaller park = Pi (D^2) / 4 (where D is the diameter of smaller circular path)
The area of bigger park = Pi [(2.5K D)^2] / 4 = 6.25(K^2) Pi (D^2)/4 or 6.25 (K^2) times the area of smaller park. The difference of the two = [6.25 (K^2) – 1] Pi (D^2) / 4. So bigger park is greater by [6.25 (K^2) – 1] times.)

Explanation

ANS: (D) (The area of the smaller park = Pi (D^2) / 4 (where D is the diameter of smaller circular path)
The area of bigger park = Pi [(2.5K D)^2] / 4 = 6.25(K^2) Pi (D^2)/4 or 6.25 (K^2) times the area of smaller park. The difference of the two = [6.25 (K^2) – 1] Pi (D^2) / 4. So bigger park is greater by [6.25 (K^2) – 1] times.)

36) What is the average age of 3 new team members who replaced 3 old ones? (1) The average age of the team decreased by 2 years. (2) The 3 replaced members were aged 31, 33 & 34 years.

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient

Statements (1) and (2) TOGETHER are NOT sufficient

(Ans: (e) Take (1), neither the total number of team members nor ages of 3 replaced members given, so (1) is not sufficient.
Now take (2), only the ages of replaced members are given. Total number of members and impact of new members on average are not given, so (2) is also not sufficient.
Now take (1) & (2) together, the information of number of members is not given without which the average of new members can not be calculated. Hence (e) is correct.)

Explanation

(Ans: (e) Take (1), neither the total number of team members nor ages of 3 replaced members given, so (1) is not sufficient.
Now take (2), only the ages of replaced members are given. Total number of members and impact of new members on average are not given, so (2) is also not sufficient.
Now take (1) & (2) together, the information of number of members is not given without which the average of new members can not be calculated. Hence (e) is correct.)

37) If 5p – s = 4q – 11, where p and q are integers and s < 4. What is the minimum value of q for which p > 77 ?

99

98

101

102

100

(Ans: (e) The equation can be rewritten as p = (4q – 11 + s)/5
Here p & q are integers but s is not integer (not mentioned), so s can take any value For q to be minimum for any particular value of p, s must be maximum. Hence we will take 4 as the maximum value of s.
The minimum value of q will be when p = 78 (>77) being integer.
Now when q = 99, then p = (389/5) And for q = 100, p = (393/5) > 78. Hence minimum value of q for which p >77 is 100.)

Explanation

(Ans: (e) The equation can be rewritten as p = (4q – 11 + s)/5
Here p & q are integers but s is not integer (not mentioned), so s can take any value For q to be minimum for any particular value of p, s must be maximum. Hence we will take 4 as the maximum value of s.
The minimum value of q will be when p = 78 (>77) being integer.
Now when q = 99, then p = (389/5) And for q = 100, p = (393/5) > 78. Hence minimum value of q for which p >77 is 100.)