ratio and proportion mcq pdf Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024
ibps rrb so officer sclae 2 3 single exam 2024 SYLLABUS WISE SUBJECTS MCQs
Number System
Lcm & Hcf
Ratio & Proportion
Percentage
Time & Work
A and B rent a pasture for 10 months; A puts in 80 cows for 7 months. In how many cows can B put for the remaining 3 months, if he pays half as much again as A ?
Answer: (d)
Let B puts in x cows
Then amount paid by B = $3/2$ × amount paid by A
∴${80 × 7}/{x × 3} = {\text"Amount paid by A"/\text"Amount paid by B"}$
= ${\text"amount paid by A"/\text"3/2 × amount paid by A"}$
⇒x = ${80 × 7 × 3}/{3 × 2}$ = 280
Hence, B puts in 280 cows
A began business with Rs.12500 and is joined afterwards by B with Rs.37500. When did B join, if the profits at the end of the year are divided equally ?
Answer: (c)
Let B join after x months of the start of the business so that B's money is invested for (12–x) months.
∴ Profit ratio is 12 × 12500 : (12 – x) × 37500
or 12 : 3(12 – x)
Since profit is equally divided. therefore
12 = 3(12 – x) or x = 8. Thus B joined after 8 months.
The number of millilitres of water added to reduce 9 ml of aftershave lotion, containing 50% alcohol, to a lotion containing 30% alcohol is
Answer: (d)
The given solution has 50% alcohol. Water which is to be added has 0% alcohol concentration.
Alcohol concentration :
∴ Water should be added in the ratio 2 : 3
∴ Quantity of water to be added = $2/3$ × 9 = 6 ml
Alternate method:
Initialy the amount of after save in solution.
= 9 × ${50}/{100}$ = 4.5ml.
let x ml of water in mixed then.
(9 + x) ${30}/{100}$ = 4.5ml (remains same)
x = 6 ml
The cost of a diamond varies directly as the square of its weight. A diamond broke into four pieces with their weights in the ratio of 1 : 2 : 3 : 4. If the loss in total value of the diamond was Rs. 70,000, what was the price of the original diamond ?
Answer: (c)
Let price of diamond be $kx^2$ where K is constant total price for 4 pieces
$Kx^2 [1 + 4 + 9 + 16] = 30 kx^2$
Price of original diamond = 100 $kx^2$
difference = $70kx^2 = 70000 or kx^2$ = 1000
original price
100 × 1000 = Rs.100000
If $1/2$ of Sunit's salary is equal to $2/5$ of Rajan's salary and their total salary is Rs. 36,000, find Rajan's salary.
Answer: (c)
Ratio of the salaries of Sumit and Rajan = $2/5 : 1/2$ = 4:5
Rajan's salary = $5/9$ × 36000 = Rs. 20000
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