Practice Excluded numbers - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A student finds the average of 10, 2 – digit numbers. If the digits of one of the numbers is interchanged, the average increases by 3.6. The difference between the digits of the 2-digit numbers is
(a)
(b)
(c)
(d)
Total increase = 3.6 × 10 = 36
∴ If the number be 10x + y,
then Number obtained after reversing the digits = 10y + x
∴ 10y + x – 10x – y = 36
⇒ 9y – 9x = 36
⇒ 9 (y – x) = 36
⇒ y – x = $36/9$ = 4
Q-2) The average of six numbers is 32. If each of the first three numbers is increased by 2 and each of the remaining three numbers is decreased by 4, then the new average is
(a)
(b)
(c)
(d)
Change = 2×3 – 3 × 4 = – 6
∴ New average = 32 -$6/6$ = 31
Q-3) If the mean of 4 observations is 20, when a constant 'C’ is added to each observation, the mean becomes 22. The value of C is :
(a)
(b)
(c)
(d)
4C = 22 × 4 – 20 × 4 = 88 – 80 = 8
⇒ C = $8/4$ = 2
Q-4) The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is :
(a)
(b)
(c)
(d)
Total sum of five numbers = 27 × 5 = 135
Total sum of four numbers = 25 × 4 = 100
∴ Required number = 135 – 100 = 35
Q-5) The average of 6 observations is 45.5. If one new observation is added to the previous observations, then the new average becomes 47. The new observation is
(a)
(b)
(c)
(d)
Let the new observation be x. Then,
${x+6×45.5}/7$= 47
⇒ x + 273 = 47 × 7 = 329
⇒ x = 329 – 273 = 56
Aliter : Using Rule 24,
Here, x = (47 – 45.5) = 1.5, n = 6
New observation = Average + x (n + 1)
= 45.5 + 1.5 (6 + 1)
= 45.5 + 10.5 = 56
Q-6) In a class, the average score of girls in an examination is 73 and that of boys is 71. The average score for the whole class is 71.8. Find the percentage of girls.
(a)
(b)
(c)
(d)
Let the number of boys be x and that of girls be y.
∴ 71x + 73y = 71.8 (x + y)
⇒ 71.8x – 71x = 73y – 71.8y
⇒ 0.8x = 1.2y
⇒ $\text"x"/ \text"y"$ = ${1.2}/{0.8}$ = $12/8$ = $3/2$
∴ $\text"x"/\text"y"$+1 = $3/2$ + 1⇒ $\text"x+y"/ \text"y"$ = $5/2$
∴ Percentage of girls
= $\text"y"/ \text"x+y"$ × 100= $2/5$ × 100= 40%
Q-7) The average of five numbers is 140. If one number is excluded, the average of the remaining four numbers is 130. The excluded number is :
(a)
(b)
(c)
(d)
Required number = 5 × 140 – 4 × 130
= 700 – 520 = 180
Aliter : Using Rule 25,
Here, x = (140 – 130) = 10, n = 5
Excluded number = Average + x (n – 1)
= 140 + 10 × 4 = 180
Q-8) The average of five numbers is 7. When three new numbers are included, the average of the eight numbers becomes 8.5. The average of the three new numbers is
(a)
(b)
(c)
(d)
Sum of the three new numbers
= 8 × 8.5 – 5 × 7 = 68 – 35 = 33
∴ Required average = $33/3$ = 11
Q-9) The average of marks of 28 students in Mathematics was 50; 8 students left the school, then this average increased by 5. What is the average of marks obtained by the students who left the school ?
(a)
(b)
(c)
(d)
Total marks of 28 students = 28× 50 = 1400
Total marks of 20 students = 20 × 55 = 1100
∴Total marks of 8 students = 1400 – 1100 = 300
∴ Average = $300/8$ = 37.5
Q-10) The average of five numbers is 7. If three new numbers would be added, then the new average comes out to be 8.5. What is the average of those three new numbers?
(a)
(b)
(c)
(d)
Sum of three new numbers = 8 × 8.5 – 7 × 5
= 68 – 35 = 33
∴ Required average = $33/3$ = 11