Test: Progression (AP And GP)- 5


15 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Test: Progression (AP And GP)- 5


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QUESTION: 1

A and B are two numbers whose AM is 25 and GM is 7. Which of the following may be a value ofA?

Solution:

(a + b)/2 = 25
a + b = 50
√ab = 7
ab = 49
Hence, A can either be 7 or 49.
So, 49 is the answer.

QUESTION: 2

In a nuclear power plant a technician is allowed an interval of maximum 100 minutes. A timerwith a bell rings at specific intervals of time such that the minutes when the timer rings are notdivisible by 2, 3, 5 and 7. The last alarm rings with a buzzer to give time for decontamination ofthe technician. How many times will the bell ring within these 100 minutes and what is the valueof the last minute when the bell rings for the last time in a 100 minute shift?

Solution:

In order to find how many times the alarm rings we need to find the number of numbers below 100
which are not divisible by 2,3, 5 or 7. This can be found by:
100 – (numbers divisible by 2) – (numbers divisible by 3 but not by 2) – (numbers divisible by
5 but not by 2 or 3) – (numbers divisible by 7 but not by 2 or 3 or 5).
Numbers divisible by 2 up to 100 would be represented by the series 2, 4, 6, 8, 10..100 Æ A total
of 50 numbers.
Numbers divisible by 3 but not by 2 up to 100 would be represented by the series 3, 9, 15, 21…
99 Æ A total of 17 numbers. Note use short cut for finding the number of number in this series :
[(last term – first term)/ common difference] + 1 = [(99 – 3)/6] + 1 = 16 + 1 = 17.
Numbers divisible by 5 but not by 2 or 3: Numbers divisible by 5 but not by 2 up to 100 would
be represented by the series 5, 15, 25, 35…95 Æ A total of 10 numbers. But from these numbers,
the numbers 15, 45 and 75 are also divisible by 3. Thus, we are left with 10 – 3 = 7 new numbers
which are divisible by 5 but not by 2 and 3.
Numbers divisible by 7, but not by 2, 3 or 5:
Numbers divisible by 7 but not by 2 upto 100 would be represented by the series 7, 21, 35, 49,
63, 77, 91 Æ A total of 7 numbers. But from these numbers we should not count 21, 35 and 63 as
they are divisible by either 3 or 5. Thus a total of 7 – 3 = 4 numbers are divisible by 7 but not by
2, 3 or 5.

QUESTION: 3

The internal angles of a plane polygon are in AP. The smallest angle is 100o and the commondifference is 10o. Find the number of sides of the polygon.

Solution:

The sum of the interior angles of a polygon are multiples of 180 and are given by (n – 1) × 180
where n is the number of sides of the polygon. Thus, the sum of interior angles of a polygon would
be a member of the series: 180, 360, 540, 720, 900, 1080, 1260
The sum of the series with first term 100 and common difference 10 would keep increasing when
we take more and more terms of the series. In order to see the number of sides of the polygon, we
should get a situation where the sum of the series represented by 100 + 110 + 120… should
become a multiple of 180. The number of sides in the polygon would then be the number of terms
in the series 100, 110, 120 at that point.
If we explore the sums of the series represented by 100 + 110 + 120…
We realize that the sum of the series becomes a multiple of 180 for 8 terms as well as for 9 terms.
It can be seen in: 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 = 1080
Or 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 + 180 = 1260.

QUESTION: 4

Each of the series 13 + 15 + 17+…. and 14 + 17 + 20+… is continued to 100 terms. Find howmany terms are identical between the two series.

Solution:

The two series till their hundredth terms are 13, 15, 17….211 and 14, 17, 20…311. The common
terms of the series would be given by the series 17, 23, 29….209. The number of terms in this
series of common terms would be 192/6 + 1 = 33. Option (d) is correct.

QUESTION: 5

A student takes a test consisting of 100 questions with differential marking is told that eachquestion after the first is worth 4 marks more than the preceding question. If the third question ofthe test is worth 9 marks. What is the maximum score that the student can obtain by attempting 98questions?

Solution:

The maximum score would be the sum of the series 9 + 13 + …. + 389 + 393 + 397 = 98 × 406/2
= 19894. Option (d) is correct.

QUESTION: 6

An equilateral triangle is drawn by joining the midpoints of the sides of another equilateraltriangle. A third equilateral triangle is drawn inside the second one joining the midpoints of thesides of the second equilateral triangle, and the process continues infinitely. Find the sum of theperimeters of all the equilateral triangles, if the side of the largest equilateral triangle is 24 units.

Solution:

The side of the first equilateral triangle being 24 units, the first perimeter is 72 units. The second
perimeter would be half of that and so on.
72, 36, 18 …

QUESTION: 7

Find the 33rd term of the sequence: 3, 8, 9, 13, 15, 18, 21, 23…

Solution:

The 33rd term of the sequence would be the 17th term of the sequence 3, 9, 15, 21 ….
The 17th term of the sequence would be 3 + 6 × 16 = 99.

QUESTION: 8

A number 20 is divided into four parts that are in AP such that the product of the first and fourth is to the product of the second and third is 2 : 3. Find the largest part.

Solution:

Since the four parts of the number are in AP and their sum is 20, the average of the four parts must
be 5. Looking at the options for the largest part, only the value of 8 fits in, as it leads us to think of
the AP 2, 4, 6, 8. In this case, the ratio of the product of the first and fourth (2 × 8) to the product
of the first and second (4 × 6) are equal. The ratio becomes 2:3.

QUESTION: 9

If a clock strikes once at one o’clock, twice at two o’clock and twelve times at 12 o’clock and again once at one o’clock and so on, how many times will the bell be struck in the course of 2days?

Solution:

In the course of 2 days the clock would strike 1 four times, 2 four times, 3 four times and so on.
Thus, the total number of times the clock would strike would be:
4 + 8 + 12 + ..48 = 26 × 12 = 312. Option (b) is correct.

QUESTION: 10

Find the sum of the integers between 1 and 200 that are multiples of 7.

Solution:

The sum of the required series of integers would be given by 7 + 14 + 21 + ….196 = 28 × 101.5 =
2842. Option (b) is correct.

QUESTION: 11

Find the sum of all odd numbers lying between 100 and 200.

Solution:

101 + 103 + 105 + … 199 = 150 × 50 = 7500

QUESTION: 12

The first and the last terms of an AP are 107 and 253. If there are five terms in this sequence, findthe sum of sequence.

Solution:

5 × average of 107 and 253 = 5 × 180 = 900. Option (c) is correct.

QUESTION: 13

If a, b, c are in GP, then log a, log b, log c are in

Solution:

If we take the values of a, b, and c as 10,100 and 1000 respectively, we get log a, log b and log c
as 1, 2 and 3 respectively. This clearly shows that the values of log a, log b and log c are in AP.

QUESTION: 14

The sum of an infinite GP whose common ratio is numerically less than 1 is 32 and the sum of thefirst two terms is 24. What will be the third term?

Solution:

Trying to plug in values we can see that the infinite sum of the GP 16, 8, 4, 2… is 32 and hence the
third term is 4.

QUESTION: 15

Find the second term of an AP if the sum of its first five even terms is equal to 15 and the sum of the first three terms is equal to –3.

Solution:

Since the sum of the first five even terms is 15, we have that the 2nd, 4th, 6th, 8th and 10th term of
the AP should add up to 15. We also need to understand that these 5 terms of the AP would also be
in an AP by themselves and hence, the value of the 6th term (being the middle term of the AP)
would be the average of 15 over 5 terms. Thus, the value of the 6th term is 3. Also, since the sum
of the first three terms of the AP is –3, we get that the 2nd term would have a value of –1. Thus, the
AP can be visualized as:
_, -1, _,_,_,3,….
Thus, it is obvious that the AP would be –2, –1, 0, 1, 2, 3. The second term is –1. Thus, option (c)
is correct.