Q1. Milk and water in two vessels A and B are in the ratios 4:3 and 2:3 respectively. In what ratio the liquids in both plates should be mixed to obtain a new mixture in vessel C, only five maintaining half milk and half water.
A.1:1
B.7:5
C.2:4
D.1:3
EXPLANATION
Option B.
The fraction of milk in A is equal to 4/(4+3)=4/7
Fraction of milk ib B= 2/(2+3)= 2/5
Let them mix in a ratio X ratio y. Such that their fraction is ½.
It can be written as –
4x/7+2y/5=x+y/2
4x/7-x/2=y/2-2y/5
x/y=7/5
The required ratio is 7:5.
Q2. There are six tickets to the theater, four of which are for seats in the front row. 3 tickets are selected at random. What is the probability that two of them are for the front row?
A.0.6
B.0.7
C.0.9
D.1/3
EXPLANATION
(A).
Out of all the six tickets, four are from the front row tickets, and the rest two are not out of forefront tickets. Two can be selected in 4C2. The remaining one ticket has to be selected in 2C1. The total ways of selecting three tickets such that two of them are front row are 4C2 *2C1
3 out of 6 tickets can be selected in 6C3 ways.
Required possibility = 4C2 * 2C1/6C3 = 0.6
Q3. When 75% of a number is added to 75, the result is same as the number. The number is-
A.150
B.300
C.100
D.450
EXPLANATION
(B).
Let the no. be x.as per question
75x/100+75=x
3x/4+75=x
x/4=75
x=4*75=300.
Q4. If five spiders can catch five flies in 5 minutes. How many flies can hundred spiders catch in 100 minutes?
A.100
B.1000
C.500
D.2000
EXPLANATION
(D) 2000.
The rate at which spiders catch the flights remains the same, hence multiplying the time frame and the number of spiders relative to the information provided gives the solution. Distributing five spiders as a group. There are 20 groups, and each group catches five spiders in five minutes. And hence, 100 spiders are caught in five minutes. And in 100 minutes=
100*(100)/5=2000 flies
M1*D1/W1=M2*D2/W2
W=2000.
Q5. The value is? 51/4 × (125)0.25
A.5
B.25
C.50
D.10
EXPLANATION
(A) 5
5 ¼ * 125 0.25
5 0.25 + 53+0.25
50.25 +0.75= 51 =5.
