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Aptitude for SBI AND RBI Mains


We bring you the complete and important daily  Quant Quiz to achieve more marks in Banking, Insurance, UPSC, SSC, CLAT, Railways and all other competitive Exams. We prepare it based on our daily current affairs.Hope you like it.

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START QUIZ


Q1. At IIM Bangalore, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 900, find the total number of students getting 50% concession, if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?
(a) 3600
(b) 2800
(c) 3200
(d) 3300
(e) None of these

Q2. A person bought some candies and sold 80% of them at the price he paid for total candies. He sold the remaining candies at 10% profit. If the total profit is 1386 Rs. then find the cost price of the all candies.
(a) 6300
(b) 6100
(c) 5700
(d) 6500
(e) 7500

Q3. S₁ is a series of five consecutive multiple of three, whose sum is 180 and S₂ is the series of four consecutive multiple of four whose second smallest number is 13 more than second highest number of S₁ series. Find the average of smallest number of S₁ series and highest number of S₂ series.
(a) 51
(b) 49
(c) 47
(d) 45
(e) 43

Q4. ‘2n’ years ago ratio of Amit’s age to Inder’s age is 5 : 4. ‘n’ years ago ratio of Inder’s age to Satish’s age is 9 : 7. Difference between present age of Amit to Satish’s present age is 12 years. Find the sum of present ages of all three if ratio of Amit’s age to Satish’s age after ‘n’ year will be 13 : 9.
(a) 81
(b) 84
(c) 87
(d) 90
(e) Cannot be determined

Q5. A student finds the average of five two digits numbers. If One number is reversed and the average is taken again then the average increase by 5.4. If all five digits are consecutive multiple of four, then find the number which is reversed?
(a) 58
(b) 36
(c) 74
(d) 48
(e) None of these

Directions (6-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer,
(a) if x<y
(b) if x≤y
(c) if x = y, or no relation can be established between x and y
(d) if x>y
(e) if x≥𝑦

Q6. I. 9x2 = 1
      II. 4y2+11y3 = 0

Q7. I. x2−5 =0
      II. 4y224y+35 = 0

Q8. I. x2−5x−14=0
      II. y2 +7y+10 = 0

Q9. I. 5x+7y=43
      II. 9x−17y=41

Q10. I. 2x2− (4+√13)x+2√13=0
         II. 10y2− (18+5√13)y+9√13=0


Solutions

S1. Ans.(d)
Sol. Let total students are x 15100×60100x+15200×40100π‘₯=900 9100x+3x100=900 12x=900×100 x=7500
Total students getting fee waiver =675+225=900
Required no. of students =(7500−900)/2=3300

S2. Ans.(a)
Sol.
Let total candies be 100 and price per candy be x
∴ total C.P = 100x
Total S.P = 100π‘₯+110/100 ×20π‘₯
ATQ, 1386=(100π‘₯+22π‘₯)−100π‘₯
required total c.p = 100× 1386/22=6300

S3. Ans.(d)
Sol.
Let S₁ is a series consists 3x – 6, 3x – 3, 3x, 3x + 3, 3x + 6
ATQ,
3x – 6 + 3x – 3 + 3x + 3x + 3 + 3x + 6 = 180
⇒ x = 12
S₁ series = 30, 33, 36, 39, 42
Second smallest no. of S₂ = 39 + 13 = 52
S₂ series = 48, 52, 56, 60
Required average= (30+60)/2=90/2=45

S4. Ans.(d)
Sol.
Let, Present age of Amit, Inder and Satish be x, y and z respectively.
ATQ,
(x –2n)/(y –2n) = 5/4
4x – 8n = 5y – 10n
5y – 4x = 2n …(i)
(y –n)/(z –n)= 9/7
7y – 7n = 9z – 9n
2n = 9z – 7y …(ii)
Equating (i) & (ii)
5y – 4x = 9z – 7y
12y = 9z + 4x
And, x – z = 12
(x+n)/(z+n)= 13/9
⇒ 9x + 9n = 13z + 13n
4n = 9x – 13z
9x – 13z = 18z – 14y
9x + 14y = 31z
Now x = 12 + z
9 (12 + z) + 14y = 31z
22z – 14y = 108
or 11z – 7y = 54 …(iii)
and 10y – 8x = 9x – 13z
17x = 10y + 13z
17 (12 + z) = 10y + 13z
204 + 17z = 10y + 13z
4z + 204 = 10y …(iv)
On solving (iii) and (iv)
y = 30, z = 24, x =36
Required sum = 30 + 24 + 36 = 90

S5. Ans.(b)
Sol.
Let the number which is reversed is “10a+b”
If it is reversed then the number becomes “10b+a”
ATQ,
10𝑏+π‘Ž−10π‘Ž−𝑏=5×5.4
9𝑏−9π‘Ž=27
⇒π‘π‘Ž=3
Numbers can be 14, 25, 36, 47, 58 and 69 but all the five digits are multiple of 4
⇒ The number should be 36

S6. Ans.(c)
Sol. π‘₯=1/3 −1/3
4𝑦2+12𝑦−𝑦−3=0
4𝑦(𝑦+3)−1(𝑦+3)=0
𝑦=14,−3
So no relation can be established
S7. Ans.(a)
Sol. π‘₯=√5,−√5
4𝑦2−14𝑦−10𝑦+35=0
2𝑦(2𝑦−7)−5(2𝑦−7)=0
𝑦=5/2, 7/2
π‘₯<𝑦
S8. Ans.(e)
Sol. π‘₯2−7π‘₯+2π‘₯−14=0
π‘₯(π‘₯−7)+2(π‘₯−7)=0
π‘₯=7,−2
𝑦2+5𝑦+2𝑦+10=0
𝑦=−2,−5
π‘₯≥y
S9. Ans.(d)
Sol. Solving these equations.
π‘₯=−3,𝑦=−4
π‘₯>𝑦
S10. Ans.(e)
Sol. 2π‘₯2−4π‘₯−√13π‘₯+2√13=0
2π‘₯(π‘₯−2)−√13(π‘₯−2)=0
π‘₯=√13/2,2
10𝑦2−18𝑦−5√13𝑦+9√13=0
2𝑦(5𝑦−9)−√13(5𝑦−9)=0
𝑦=√13/2,95
π‘₯≥𝑦