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Question 1 of 5
1. Question
Two persons A and B start simultaneously from two places c km apart and walk in the same direction. If A travels at the rate of p km/h and B travels at the rate of q km/h, then A has travelled before he overtakes B a distance of
Correct
Ans: b
Let A and B will meet after t h at point E.
Distance travelled by A = pt h
[because, distance = speed * time]
and distance travelled by B = qt h
According to the question,
pt = qt +C ⇒ pt – qt = c
t(p-q) = c
t=(c/p-q) …………………………………………………………(1)
∴Distance travelled by A = pt = (pc/(p-q)) km [from Eq.(1)]
Incorrect
Ans: b
Let A and B will meet after t h at point E.
Distance travelled by A = pt h
[because, distance = speed * time]
and distance travelled by B = qt h
According to the question,
pt = qt +C ⇒ pt – qt = c
t(p-q) = c
t=(c/p-q) …………………………………………………………(1)
∴Distance travelled by A = pt = (pc/(p-q)) km [from Eq.(1)]
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Question 2 of 5
2. Question
As per agreement with a bank, a businessman to refund a loan in some equal instalments without interest. After paying 18 instalments, he found that 60% of his loan was refunded. How many instalments were there in the agreement?
Correct
Ans: c
In this question, the loan is refunded with equal installment and without interest. Initially the man had to pay x installations. Now, after paying 18 installments, 60 % of loan iş refunded .
60 % of x = 18 = 60/100* X= 18
18 x 100/60 = 30
There were 30 installations in the agreement .
Incorrect
Ans: c
In this question, the loan is refunded with equal installment and without interest. Initially the man had to pay x installations. Now, after paying 18 installments, 60 % of loan iş refunded .
60 % of x = 18 = 60/100* X= 18
18 x 100/60 = 30
There were 30 installations in the agreement .
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Question 3 of 5
3. Question
Consider the following in respect of the numbers , √2, 3 √3 and 6 √6
3. In a five digit number, last digit is three times the first one, 3rd digit is 3 less than the second,4th digit is 4 more than the second one . Find the number.
Correct
Ans. A
We are given a 5 digit number
Let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y – 3’
and 4th digit is ‘Y + 4’
then the no is ‘(X)(Y)(Y – 3)(Y + 4)(3X)’
from the above we can say 3X <= 9
so X<=3 and any of the digit in the number is <= 9Also Y–3 ≥ 0 and Y+4 ≤ 9
Y ≥ 3 and Y ≤ 5 i.e. Y = 3,4 or 5
for x = 1, all the conditions won’t be satisfied.For x = 2,
Let Y = 5
The number is 25296Short trick
Since the relation between the digits are given, let us go through the option and see if any one of them satisfies the condition.
Option 1: 25296
6 = 3 × 2, 2 = 5 – 3, 9 = 5 + 4
Since all the conditions are satisfied, option 1 is the solution.
Incorrect
Ans. A
We are given a 5 digit number
Let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y – 3’
and 4th digit is ‘Y + 4’
then the no is ‘(X)(Y)(Y – 3)(Y + 4)(3X)’
from the above we can say 3X <= 9
so X<=3 and any of the digit in the number is <= 9Also Y–3 ≥ 0 and Y+4 ≤ 9
Y ≥ 3 and Y ≤ 5 i.e. Y = 3,4 or 5
for x = 1, all the conditions won’t be satisfied.For x = 2,
Let Y = 5
The number is 25296Short trick
Since the relation between the digits are given, let us go through the option and see if any one of them satisfies the condition.
Option 1: 25296
6 = 3 × 2, 2 = 5 – 3, 9 = 5 + 4
Since all the conditions are satisfied, option 1 is the solution.
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Question 4 of 5
4. Question
What is the harmonic mean of 10, 20, 25, 40 and 50?
Correct
Ans: d
Let the numbers be a1 = 10, a2 = 20, a3=25, a4 = 40 and a5 = 50
∴Harmonic mean = (Number of observations)/[(1/a1)+(1/a2)+(1/a3)+(1/a4)+(1/a5)]
Incorrect
Ans: d
Let the numbers be a1 = 10, a2 = 20, a3=25, a4 = 40 and a5 = 50
∴Harmonic mean = (Number of observations)/[(1/a1)+(1/a2)+(1/a3)+(1/a4)+(1/a5)]
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Question 5 of 5
5. Question
The digit in the unit’s place of the resulting number of the expression (234)100 + (234)101, is
Correct
Ans: d
We have, (234)100+ (234)101 (234)0= 235(234)100
We know that square of any number having 4 at unit place is a number in which 6 at unit place. Any exponent of a number 6 at unit place is always 6 is at unit place.
Because (235)(…..6)=…30
Resulting number have 0 at unit place.
Incorrect
Ans: d
We have, (234)100+ (234)101 (234)0= 235(234)100
We know that square of any number having 4 at unit place is a number in which 6 at unit place. Any exponent of a number 6 at unit place is always 6 is at unit place.
Because (235)(…..6)=…30
Resulting number have 0 at unit place.
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