[CSP] Test PT19ST21 – CSAT

#1. A person travels 12km due North, then 15km due East, after that 15km due West and then 18km due South. How far is he from the starting point?

(a) 6 km

(b) 12km

(c) 33 km

(d) 60 km

#2. At present, the age of a father is equal to sum of ages of his six children. After 10 years, his age will be half of the sum of ages of his six children. What is his present age?

(a) 40 years

(b) 45 years

(c) 50 years

(d) 55 years

#3. What will be the 100th alphabet in the following sequence?

a, b, b, c, c, c, d, d, d, d,……

(a) m

(b) n

(c) p

(d) q

#4. If a number is added to its reciprocal, the resulting number is 25% more than the original number. The original number is

(a) 2

(b) 4

(c) 6

(d) 8

#5. In the wake of demonetization, a merchant exchanged a certain amount, all in the form of old Rs. 1000/- denomination notes with the new Rs. 500/- and Rs. 2000/- denomination notes. If the number of Rs. 500/- and Rs. 2000/- denomination notes received by him for the exchange was same then what is the minimum amount he must have exchanged?

(a) Rs. 25,000/-

(b) Rs. 5,000/-

(c) Rs. 20,000/-

(d) Rs. 10,000/-

#6. A bag contains a certain number of balls each of which is either red or green. If two red balls in the bag are replaced with green balls, the ratio of number of red to green balls in the bag becomes 1:2, whereas if three green balls are replaced with red balls the same ratio becomes 2:1. Which of the following statements is not correct?

(a) There are more number of red balls in the bag than the green balls

(b) There are even number of green balls in the bag

(c) In all there are odd number of balls in the bag

(d) There the one more green balls in the bag than the red balls

#7. During an experiment, a sample of a radio- active material with half-life of 4 hours was placed in a container. The sample was taken out after 48 hours and weighed 2 gms. What was the weight of the sample when it was placed in the container?

(a) 4096 gms

(b) 2048 gms

(c) 1024 gms

(d) 8192 gms

#8. A parking lot contains 80 vehicles. Each vehicle is either a car or a truck and each vehicle is either red or green. If 35 are red, 60 vehicles are cars and there are 9 green trucks then how many red cars are there?

(a) 30

(b) 24

(c) 26

(d) 28

#9. A man rows downstream 32 km and 14 km upstream, and he takes 6 hours to cover each distance. What is the speed of the current?

(a) 0.5 km/hr

(b) 1 km/hr

(c) 1.5 km/hr

(d) 2 km/hr

#10. How many times the digit 5 is printed, if a book containing 250 pages is to be numbered from 1 to 250?

(a) 48

(b) 42

(c) 50

(d) 46

Solutions:

Ans 1: (a)


Let S be the starting Point. The person travels 12km north and reached N. The person travels 15 km due east and reaches E. Then the person travels 15km due west and consequently reaches Point N again. Then the person travels 18 km south and consequently passes through s, the starting point, and travels further south 6km to reach final point F.

Ans 2: (a)

Let the present age of the father be x (which is equal sum of ages of his six children at present). After 10 years, father’s age = x + 10 and sum of ages of his six children = x + 60 (10 years added to the age of each child)

It is given that x + 10 = (1/2) (x + 60)
Solving for x we get x = 40 years

Ans 3: (b)

a appears 1 time, b appears 2 times, c appears 3 times and so on.

By the time we reach m (the 13th alphabet) we would have written 1 + 2 + 3 + 4 +…… + 13 = 91 alphabets. Then we would write n 14 times and thus 100th alphabet would be n.

Ans 4: (a)

Let x be the number then as per the condition given
x + (1/x) = x + (25% of x)
This implies that 1/ x = 25x/100
and thus x = 2.

Alternatively :
Use the option: Let us begin with first option ‘2’.
2 + (1/2) = 2.5 and 0.5 is 25 percent of 2. Thus 2 is the answer.

Ans 5: (b)

The merchant received equal number of Rs. 500 and Rs. 2000 notes. Let the number of each be x. Thus he received a sum of Rs. 500X + 2000x = 2500x in exchange.

Now the amount that he had deposited was all in the Rs. 1000 denomination….Condition 1
If x = 1, the amount was Rs. 2500 which is not possible because of Condition 1
If x = 2, then the deposited amount was Rs. 5000 which satisfies condition 1. Thus the minimum amount satisfying the given conditions is Rs. 5000.

Ans 6: (a)

Let number of red and green balls be R and G respectively.
If two red balls are replaced with green balls, the number of red and green balls becomes R-2 and G + 2. Since the ratio of red to green balls is now 1:2 thus
(R-2) / (G+ 2)= 1/2
=> 2R-4 = G + 2
=> 2R – G = 6 …….. Condition 1
Now if three green balls are replaced with red balls, the number of red and green balls becomes R+3 and G -3. Since the ratio of red to green balls is now 2:1 thus
(R+3) / (G – 3)= 2/1
=> R + 3 = 2G – 6
=> R – 2G = -9 …….. Condition 2

Solving Condition 1 and 2 for R and G we get, R = 7 and G = 8.

Now if we check the option statements given we find that only option statement (a) is not correct.

Ans 7: (d)

Let the amount of sample placed in the container was x.
After 4 hours (i.e. One half life), the amount left shall be x/2.
After another 4 hours ( i.e. One more half life), the amount left shall be x/4 and so on.
After 48 hours i.e. 12 half lives, the amount left shall be x/(2 raised to the power 12) = 2gm as given.
=> x = (2 raised to the power 12) × 2 gms = 2048 gms

Ans 8: (b)

The information can be formulated as under:


Now, total vehicles = 80, cars = 60 and thus trucks = 20.
Total red = 35, thus green = 80 – 35 = 45
Total green = 45, thus green cars = 45 – 9 = 36
Total cars = 60, total green cars = 36, thus total red cars = 60 – 36 = 24

Ans 9: (c)

Rate downstream movement, D = 32/6 kmph

Rate upstream movement, U= 14/6 kmph

Velocity of current = 1/2 (D – U) = 1/2 (32/6 – 14/6) kmph = 3/2 kmph = 1.5 kmph

Ans 10: (d)

5,15,25,35,45,50,51,52,53,54,55,56,57,58,59,65,75,85,95,105,115,125,135,145, 150,151,151,153,154,155,156,157,158,159,165,175,185,195,205,215,225,235, 245,250

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