1.
The population of a city A which is 68000 decreases at the rate of 1200/year. Population of city B which is 42000, increases at the rate of 800 per year. Find in how many years, the population of cities A and B are equal?
Answer: Option 'B'
We have to find the population of cities A and B after x years.
Step 1: Population of city A = 68000, decreases at the rate of 1200/year
68000 – 1200x
Step 2: Population of city B = 42000, increases at the rate of 800/year
42000 + 800x
Step 3: Find after how many population of cities A and B are equal.
Population of city A = Population of city B
68000 – 1200x = 42000 + 800x
68000 – 42000 = 1200x + 800x
26000 = 2000x
x = 13
2.
853 + ? ÷ 17 = 1000
Answer: Option 'A'
853 + ?/17 = 1000
?/17 = 1000 – 853 = 147
? = 17 × 147 = 2499
3.
(764 × ?) ÷ 250 = 382
Answer: Option 'B'
(764 × ?)/250 = 382
? = (382 × 250)/764 = 125
4.
√? + 416 = (60% of 920) – 110
Answer: Option 'A'
√? + 416 = (60% of 920) – 110
√? = (50% of 920 + 10% of 920) – 110 – 416
√? = (460 + 92) –526
√? = 552 – 526 = 26
⇒ ? = 262 = 676
5.
1035 ÷ [(3/4) of (71 + 65) – 15 ¾] = ?
Answer: Option 'A'
1035 ÷ [(3/4)× 136 – 63/4] = x
X = 1035 ÷ [102 – (63/4)]
X = 1035 ÷ [(408 – 63)/4]
X = 1035 ÷ (345/4)
X = 1035 × 4/345
X = 12
6.
Simplify: y – [y – (x + y) – {y –(y – x – y )} + 2x]
Answer: Option 'C'
Follow the rule of ‘VBODMAS’
Step 1: Solve the expression below bar
y-[y-(x+y)-{y-(y-x+y) }+2x]
Step 2: Solve () bracket
y-[y-(x+y)-{y-(2y-x) }+2x]
y-[-x-{y-2y+x}+2x]
Step 2: Solve {} bracket
y-[-x-{y-2y+x+2x}]
y-[-x-{-y+3x}]
y-[-x+y-3x]
Step 3: Solve [ ] bracket
y+4x-y
4x
Therefore, correct answer is 4x i.e option (c)
7.
(45)2 + (21)2 = (?)2 + 257
Answer: Option 'C'
(45)2 + (21)2 = (?)2 + 257
or, (?)2 = (45)2 + (21)2 – 257
or, (?)2 = 2025 + 441 – 257 = 2209.
or, ? = 2209 = 47
8.
52 + 132 – 112 = (?)3 – 52
Answer: Option 'C'
From the given equation
⇒ 25 + 169 – 121 = ?3 – 52
⇒ 25 + 48 + 52 = ?3
⇒ ?3 = 125
⇒ ?3 = (5)3 ⇒ ? = 5.
Hence, option C is correct.
9.
98643 – 21748 = 51212 + ?
Answer: Option 'A'
Given eqn,
98643 – 21748 = 51212 + ?
⇒ ? = 98643 – 21748 – 51212
⇒ ? = 25683
10.
A contractor pays Rs. 20 to a worker for each day and the worker forfeits Rs. 10 for each day if he is idle. At the end of 60 days, the worker gets Rs. 300. Find for how many days the worker was idle?
Answer: Option 'B'
Step 1: Number of days worked by the worker = 60 and he remained idle for x days. Therefore, number of days worked = (60 – x)
Step 2: Each day he was getting paid Rs. 20. Therefore, the payment received for working days = (60 – x) 20
Step 3: After subtracting the amount which he forfeited, he receives Rs. 300.
Therefore,
(60 – x) 20 – 10x = 300
1200 – 20x – 10x = 300
900 = 30x
x = 30 days