GRE PAPER- II
1. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. xy = r, xz = r2, yz = r3 x + y + z = 13 and x2 + y2 + z2 = 91 Column A Column B z/y y/x (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
2 2 2 2 Solution. (c) ( x + y + z) = x + y + z + 2( xy + yz + zx )
⇒ 169 = 91 + 2(r + r 2 + r 3 ) ⇒ r + r 2 + r 3 = 39 ⇒ r 3 + r 2 + r − 39 = 0 ⇒ ( r − 3)(r 2 + 4r + 13) = 0 ⇒ r = 3
Column A :
z xz r 2 = = =r=3 y xy r
y yz r 3 = = =r=3 2 Column B : x xz r
2.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. The entrance of a building is in the shape of a triangle placed over a rectangle as shown in the figure.
5m h 6m 4m 5m
Column A Column B The height of the top of the entrance from 2 times the breadth of the rectangle the ground (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given Solution. (c) Column B : Column A : 52 = 32 + h2 ⇒ h = 4 m. ∴ required height = 4 + 4 = 8 m. Breadth of rectangle = 4 m.
3.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Three unbiased coins are tossed. Column A Column B The probability of getting at least 2 heads The probability of getting atmost two heads (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given S = {HHH, HHT, HTT, TTT, HTH, TTH, THT, THH} Favorable cases = {HHH, HHT, HTH, THH}
4 1 = 8 2. ∴ required Prob. =
Solution. (b)Column A :
Column B :
S = {HHH, HHT, HTT, TTT, HTH, TTH, THT, THH} Favorable cases = {HHT, HTT, TTT, HTH, TTH, THT, THH}
7 ∴ required Prob. = 8 .
4.
The ratio of milk and water in 55 litres of adulterated milk is 7 : 4. How much water must be added to make the mixture's ratio 7 : 6? (a) 5 litres (b) 10 litres (c) 15 litres (d) 25 litres (e) 30 litres
Milk = 55 × 7 = 35 litres 11
Solution.
(b)
Water = 55 × 4 = 20 litres 11
Let x litres of water is added to make the ratio of milk and water to 7 : 6.
35 7 = 20 + x 6 Now,
or 140 + 7x = 210 or 7x = 70 or x = 10 litres 5. A light blinks such that it is on for 1.5 sec and off for 1 sec. If it is switched on at 06 : 43 : 20 am and switched off at 11 : 21 : 13 p.m., on the same day, then in what position is the light when switched off? (a) about to switch on (b) on (c) about to switch off (d) off (e) Neither on nor off (b) Time for a complete cycle = 2.5 sec. Then complete cycles are upto 11 : 21 : 12.5 p.m. At this time light blinks on and it remains upto 11 : 21 : 14 p.m. So it was on when switched off.
Solution.
6.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B The smallest three digit prime number 103 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
Solution. (b) Column A : 100 is not prime. 101 is prime and thus it is the smallest three digit prime number.
7.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. 15 toys cost $234. Column A Column B 35 toys cost 550 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
Toys 15 (b) Column A : 35 Cost 234 x
Solution.
It's direct proportion. Therefore 15:35::234:x ⇒ 15 x x = 35 x 234 ⇒ x = (35 x 234)/15 = 546
8.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B The remainder when x2 + 10x + 30 is The remainder when x2 + 5x – 6 is divided by x + 4 divided by x + 5 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
Solution. (a) Column A : Remainder = (–4)2 + 10(–4) + 30 = 6. Column B : Remainder = (–5)2 + 5(–5) – 6 = – 6. 9. Three men start together to travel the same way around a circular track of 11 km. Their speeds are 4, 5½ and 8 km per hour, respectively. When will they meet at the starting point? (a) 22 hrs (b) 12 hrs (c) 11 hrs (d) 44 hrs (e) 33 hrs
11 11 11 , , Solution. (a) Their timings are 4 5½ 8 . 11 2 11 LCM of 11, 2, 11 22 , , = = = 22 hours. HCF of 4, 1, 8 1 ∴ the L.C.M. of 4 1 8 2
10.
A scooter before overhauling requires 3 hour service time every 45 days, while after overhauling it requires
2 3 hour service time every 60 days. What fraction of pre-overhauling service time is saved in the latter case? 4 1 3 1 3 (a) (b) 3 (c) 4 (d) 4
(e) Solution.
4 9
(d)
L.C.M. of 45 and 60 = 180 days
2 ⎛ 180 ⎞ 8 ×⎜ ⎟= 3 ⎝ 45 ⎠ 3
In 180 days, pre-overhauling service time =
In 180 days, after overhauling service time =
8 −2 Time saved 1 = 3 = 8 Pre - overhauling time 4 3
2 ⎛ 180 ⎞ ×⎜ ⎟=2 3 ⎝ 60 ⎠
11.
A monkey ascends a greased pole 12 metres high. He ascends 2 metres in first minute and slips down 1 metre in the alternate minute. In which minute, he reaches the top? (a) 20th (b) 21th (c) 22th (d) 23th (e) 24th (b) In two minutes, the monkey ascends by a height of (2 – 1) = 1 metres. The monkey ascends 10 metres in 20 minutes ⇒ The monkey ascends 12 metres in 21 minutes
Solution.
Directions for questions 12 and 13 : The pie-chart provided below gives the distribution of land (in a village) under various food crops. Study the pie-chart carefully and answer the questions that follow: Distribution of areas (in acres) under various food crops
Wheat Rice Barley 36° Jowar 18° 18° 45° Bajra Maize 72° 72° 99° Others
12.
If the total area under jowar was 1.5 million acres, then what was the area (in million acres) under rice ? (a) 6 (b) 7.5 (c) 9 (d) 4.5 (e) 8.5 (a) The area under any of the food crops is proportional to the central angle corresponding to that crop. Let, the area under rice production be x million acres. Then, 18 : 72 = 1.5 : x Thus, the area under rice production = 6 million acres.
⇒
Solution.
⎛ 72 × 1.5 ⎞ x=⎜ ⎟=6 ⎝ 18 ⎠
13.
Which combination of three crops contribute to 50% of the total area under the food crops? (a) Wheat, Barley and Jowar (b) Rice, Wheat and Jowar (c) Rice, Wheat and Barley (d) Bajra, Maize and Rice (e) Jowar, Bajra and Wheat (c) The total of the central angles corresponding to the three crops which cover 50% of the total area, should be 180°. Now, the total of the central angles for the given combinations are: (a) Wheat, Barley and Jowar = (72° + 36° + 18°) = 126° (b) Rice, Wheat and Jowar = (72° + 72° + 18°) = 162° (c) Rice, Wheat and Barley = (72° + 72° + 36°) = 180°
Solution.
(d) Bajra, Maize and Rice = (18° + 45° + 72°) = 135° Clearly, (c) is the required combination. 14. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. A and B together can complete a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. Column A Column B The time taken by A and C to finish the same The time taken by A and B work. (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
1 (c) (A + B)'s one day work = 8 th part of total work 1 12 th part of total work (B + C)'s one day work = 1 (A+B+C)'s one day work = 6 th part of total work
Solution.
1 1 1 − = 6 12 12 th part of total work ⇒ A's one day work = 1 1 1 − = ⇒ B's one day work = 8 12 24 th part of total work
1
⇒ C's one day work = 12 24 24 th part of total work Column A : (A + C)'s one day work = . ∴ A and C can complete the work in 8 days. Column B : (A + B)'s one day work = . ∴ A and B can complete the work in 8 days. 15. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. A polygon has 35 diagonals. Column A Column B The number of sides of polygon 12 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given Column A : Let the sides of polygon = n ∴ number of diagonals = 35 = nC2 – n
⇒ 35 = n! n(n − 1) − n ⇒ 35 = −n 2!(n − 2)! 2
⇒ n 2 − 3n − 70 = 0 ⇒ n = 10.
−
1
=
1
Solution. (b)
16.
If a car moves from A to B at a speed of 60 km/h and returns back to A at a speed of 40 km/h, then find its average speed during the journey. (a) 50 km/h (b) 48 km/h (c) 55 km/h (d) 45 km/h (e) 52 km/h
Solution.
(b) Here both distances are same So, using the formula for average speed, we have
V= 2V1 V2 V1 + V2
2 × 60 × 40 100 = = 48 km/h.
17.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. In a certain school 20% of students are below 10 years of age. The number of students above 10 years of age is 2/3 of the number of students of 10 years age which is 48. Column A Column B Total number of students in the school 80 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
Solution. (a) Let total number of students = N Number of students of or above 10 years = (100 – 20)% of N = 80% of N
2
Now, Number of students above 10 years = 3 of Number of students of 10 years
= 2 2 of 48 = × 48 = 32. 3 3 ⇒ 80% of N = 32 + 48 ⇒ 80 × N = 80 ⇒ N = 100. 100
18.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B (a) (b) (c) (d) 9 quantity in column A is greater quantity in column B is greater both quantities are equal the relationship cannot be determined from the information given
( 243) 0.8 = ( 243) 0.8− 0.4 = ( 243) 0.4 = (35 ) 0.4 = 32 = 9
(243)0.8 ÷ (243)0.4
Solution. 19.
0.4 (c) (243)
In this figure below, FED is a straight line, ∠D is equal to
B F A 70° 80°
E C
D
(a) 30° Solution.
(b) 60°
(c) 40°
(d) 50°
(a) Given, FED is a straight line. ⇒ In ΔBFD, ∠F + ∠B + ∠D = 180º or ∠D = 180º – (70º + 80º) = 180º – 150º = 30º
20.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B Distance travelled by a car of speed Distance travelled by a car of speed 40 km/h in 2 hours 30 km/h in 3 hours (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
Solution. (b) Column A : Distance = Speed × Time = 40 × 2 = 80 Km. Column B : Distance = 30 × 3 = 90 Km. 21. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer.
Illiterate Female 8%
Literate Female 33%
Literate Male 35%
Illiterate Male 24% N = 2,50,000
Percentage of Male, Female, Literate and Illiterate. Column A Column B The difference between number of Literate 5,000 Male and Literate Female (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given
2 × 250000 = 5000. 100 (c) Column A : Required difference = (35 – 33)% of 250000 = 2% of 250000 =
Solution. 22.
A man earns $20 on the first day and spends $15 on the next day. He again earns $20 on the third day and spends $15 on the fourth day. If he continues to save like this, how soon will he have $60 in hand? (a) On 17th day (b) On 27th day (c) On 30th day (d) On 40th day (e) On 45th day (a) Money earned in 2 days = $(20 – 15) = $5.
⎛5 ⎞ ⎜ ×16 ⎟ ⎝2 ⎠ = $40. Money earned in 16 days = $
Solution.
On 17th day, money in hand = $(40 + 20) = $60. 23. DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer.
E 50° 85° A O 15° B D C
AB is diameter
Column A
∠ECA
Column B 20°
(a) (b) (c) (d)
quantity in column A is greater quantity in column B is greater both quantities are equal the relationship cannot be determined from the information given
∠EOD = 180° − (85° + 15°) = 80°
Solution. (a)
∴ ∠ODE = 180° − (50° + 80°) = 50° ⇒ ∠ODC = 180° − 50° = 130° ⇒ ∠OCD = 180° − (130° + 15°) = 35°
Directions for questions 24 and 25 : The bar graph given below shows the foreign exchange reserves of a country (in million US $) from 1991-92 to 1998-99. Answer the question based on this graph.
6000 5000 Foreign exchange reserves (in million US $) 4000 3000 2000 1000 2640 3720 3360 2520 3120 3120 5040 4320
1992-93
1993-94
1994-95
1995-96
1996-97
1997-98
1991-92
Years
24.
The foreign exchange reserves in 1997-98 was how many times that in 1994-95 ? (a) 0.7 (b) 1.2 (c) 1.4 (d) 1.5 (e) 1.8
5040 = 1.5
Solution. 25.
(d)
Required ratio = 3360
.
The foreign exchange reserves in 1996-97 were approximately what percent of the average foreign exchange reserves over the period under review? (a) 95% (b) 110% (c) 115% (d) 125% (e) 140%
1998-99
Solution.
(d) Average foreign exchange reserves over the given period
⎡1 ⎤ ⎢ 8 × (2640 + 3720 + 2520 + 3360 + 3120 + 4320 + 5040 + 3120⎥ ⎣ ⎦ = million US $
= 3480 million US $ Foreign exchange reserves in 1996-97 = 4320 million US $
⎛ 4320 ⎞ × 100⎟% = 124.14% ≈ 125% ⎜ ⎝ 3480 ⎠ Required Percentage =
26.
DIRECTIONS: The question consists of two quantities. One in Column A and one in Column B, you are to compare the two quantities and then give the answer. Column A Column B The average of all the numbers between 25 6 and 34, which are divisible by 5 (a) quantity in column A is greater (b) quantity in column B is greater (c) both quantities are equal (d) the relationship cannot be determined from the information given Column A : The numbers are 10, 15, 20, 25, 30.
10 + 15 + 20 + 25 + 30 100 = = 20. 5 5 ∴ Required Average =
Solution. (b)
27.
A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. If no worker was withdrawn at any stage, the group would have finished the job in two-thirds of the time. How many workers were there in the group? (a) 2 (b) 3 (c) 5 (d) 10 (e) 6 (b) Let a be the amount of work done by 1 man in 1 day. Also, let n be the number of days required to do the job. On the first day n men worked, second day (n – 1) men ................ (n – 1)th day, n – (n – 2) = 2 men and on the nth day 1 man worked to complete the work. The amount of work done by these men is na + (n – 1)a + (n – 2)a +............2a + a = 1, i.e. an (n + 1) = 2 .........(i) If no worker was withdrawn on any day, n men working for 2/3n days complete the work. i.e. na × 2/3n = 1 or 2n2a = 3 ........(ii) (i)/(ii) gives n(n + 1)/2n2 = 2/3 or n + 1/n = 4/3 giving 3n + 3 = 4n or n = 3 ∴ There were three workers in the group. Alternative Method : From the problem itself we have the clue that the group of workers would have done the job in 2/3 days. This leads us to the alternative choice b which is 3 workers complete the job in 2 days.
Solution.
28.
2x − 3 5x − 2 , then find the value of f(y). If y = f (x) =
(a) x (e) None of these
(b) 2x
(c)
x 2
(d)
5x 3
Solution.
(a)
y = f (x) =
2x − 3 5x − 2
⎛ 2x − 3 ⎞ 2⎜ ⎟−3 2y − 3 5x − 2 ⎠ = ⎝ f ( y) = 5y − 3 ⎛ 2x − 3 ⎞ 4x − 6 − 15x + 6 −11x 5⎜ ⎟−2 = = =x ⎝ 5x − 2 ⎠ 10x − 15 − 10x + 4 −11