CGI Group Aptitude Test Paper

Posted on :21-03-2016
Q1. The retail price of a water geyser is Rs. 1,265. If the manufacturer gains 10 %, the wholesale dealer gains 15 % and the retailer gains 25 %, then the cost of the product is:

A) 800
B) 900
C) 700
D) 600

ANS: A

Explanation:

C.P = 1265*100*100*100/110/115/125
C.P = 800

Q2. What percent of selling price would be 34 % of cost price if gross profit is 26 % of the selling price?

A) 17.16
B) 74
C) 25.16
D) 88.40

ANS: C

Explanation:

X% of SP = 34% of CP
Also, P = 26% of SP
SP - CP = 0.25(SP)
CP = 0.74(SP)
Now, 34/100*74
X = 25.16

Q3. The tax on a commodity is diminished by 10 % and its consumption increased by 10 %. The effect on the revenue derived from it changes by K %. Find the value of K.

A) 1.
B) -2
C) -1
D) 2

ANS: C

Explanation:

Directly using the formula, when a value is increased by R% and then decreased by R%, then net there is ( R^2)/100 decrease. Putting R = 10, we get 1% decrease.

Q4. Ratio of Ashoks age to Pradeeps age is 4 : 3. Ashok will be 26 years old after 6 years. How old is Pradeep now?

A) 18
B) 21
C) 15
D) 24

ANS: C

Explanation:

Given A/p= 4/3 Also A = 26 after 6 years, so his present age = 20years, Substituting we get P = 15 years.

Q5. The incomes of Chanda and Kim are in the ratio 9 : 4 and their expenditures are in the ratio 7 : 3. If each saves Rs. 2,000, then Chandas expenditure is

A) 60000
B) 80000
C) 70000
D) None of these

ANS: C

Explanation:

Let the incomes of Chanda and Kim be 9x and expenditures be 7y and 3y respectively. Since = Income – Expenditure, we get 9x – 7y = 2000 and 4x – 3y = 2000. Solving, we get, x = 8000 and y = 10000. So Chandas expenditure = 7y = 7 × 10000 = Rs. 70,000.

Q6. A student purchased a computer system and a colour printer. If he sold the computer system at 10 % loss and the colour printer at 20 % gain, he would not lose anything. But if he sells the computer system at 5 % gain and the colour printer at 15 % loss, he would lose Rs. 800 in the bargain. How much did he pay for the colour printer?

A) 8000
B) 16000
C) 9000
D) 5334

ANS: B

Explanation:

Let C and P be the cost price of Computer and Printer respectively.
So CP = C + P, Case I,
SP = 0.9C + 1.2P.
C = 2P.Since he did not lose anything C + P = 0.9C + 1.2P
Case II, SP = 1.05C + 0.85P
Since there was the loss of Rs. 800
Rs.800 = C + P – 1.05C – 0.85P
80000 = 15P – 5C
Using equation from Case I, we get P = Rs.16000.

Q7. X and Y entered into partnership with Rs. 700 and Rs. 600 respectively. After 3 months X withdrew 2/7 of his stock but after 3 months, he puts back 3/5 of what he had withdrawn. The profit at the end of the year is Rs. 726. How much of this should X receive?

A) 336
B) 366
C) 633
D) 663

ANS: B

Explanation:

Xs profit : Ys profit
= 700 × 3 + 500 × 3 + 620 × 6 : 600 × 12
= 2,100 + 1,500 + 3,720 : 7,200
= 7,320 : 7,200
= 61 : 60
Xs share in the profit = 726/(60+61)=366

Q8. A man sitting in a train travelling at the rate of 50 km/hr observes that it takes 9 sec for a goods train travelling in the opposite direction to pass him. If the goods train is 187.5 m long, find its speed.

A) 25
B) 45
C) 35
D) 36

ANS: A

Explanation:

Let required speed be x.
So,187.5/{ (x+50)*5/18} =9

Q9. A runs 5/3 times as fast as B. If A gives B a start of 80m, how far must the winning post be, so that A and B might reach it at the same time? <

A) 200
B) 300
C) 270
D) 160

ANS: A

Explanation:

Ratio of speeds of A : B = 5 : 3, So If A runs 5, B runs 3. So difference in distance = 2. So if difference is 2, winning post is 5m. Hence if difference is 2.5*80, winning post is Hence 1st option.

Q10. A team of workers was employed by a contractor who undertook to finish 360 pieces of an article in a certain number of days. Making four more pieces per day than was planned, they could complete the job a day ahead of schedule. How many days did they take to complete the job?

A) 8
B) 9
C) 10
D) 12

ANS: B

Explanation:

Days taken in the general scenario = 360/N;
Days taken when 4 articles are prepared extra per day = 360/N + 4;
The difference in the day is one, therefore;
360/n - 360/n+4 =1
N2 + 4N – 1440 = 0;
N = 36, i.e. number of item prepared in general scenario is 36, and where 4 articles prepared extra is 40. Therefore no of days taken to complete the job = 360/40 = 9.

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