Gauss Contest (NMTC PRIMARY 2018 - V and VI Grades) - Stage I- Problems and Solution

Part A

Problem 1

Observe the following sequence. What is the 100th term?
$$
7,8,1,0,0,1,0,1,1,0,2,1,0,3, \ldots \ldots \ldots
$$
(A) 1
(B) 0
(C) 2
(D) 3

Problem 2

A number is multiplied by 2 then by $\frac{1}{3}$, then by 4 , then by $\frac{1}{5}$ then by 6 and finally by $\frac{1}{7}$. The answer is 16 . Then the number is
(A) odd
(B) even
(C) Square
(D) a cube

Problem 3

Samrud bought a t- shirt for Rs.250. His friend Shlok wanted by buy it. Samrud wants to have a $10 \%$ profit on that. The selling price is (in rupees)
(A) 280
(B) 278
(C) 276
(D) 275

Problem 4

The value of $1+21+4161+81-11-31-51-71-91$ is
(A) -50
(B) 50
(C) 100
(D) -100

Problem 5

In the adjoining figure what portion of the figure is shaded ?

(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) $\frac{3}{10}$

Problem 6

The sum of the numbers in the three brackets ( ) is
$$\frac{()}{24}=\frac{20}{()}=\frac{24}{18}=\frac{4}{()}$$
(A) 60
(B) 55
(C) 50
(D) 45

Problem 7

A is the smallest three digit number which leaves a remainder 2 when divided by $17 . B$ is the smallest three digit number which leaves remainder 7 . When divided by 12 . Then $A+B$ is
(A) 205
(B) 312
(C) 215
(D) 207

Problem 8

A square of side 3 cm in cut into 9 equal squares. Another square of side 4 cm is cut into 16 equal squares. Saket made a bigger square using all the smaller square bits. The length of the side of the bigger square is (in cm)

(A) 7
(B) 6
(C) 5
(D) 8

Problem 9

A contractor constructed a big hall, rectangular in shape, with length 32 meters and breadth 18 meters. He wanted to buy 1 meter by 1 meter tiles. But in the shop 3 meter by 2 meter tiles only were available. How many tiles he has to buy for tilting the floor?

(A) 48
(B) 96
(C) 120
(D) 126

Problem 10

The fraction to be added to $2 \frac{1}{3}$ to get the fraction $4 \frac{4}{7}$ is
(A) $2 \frac{1}{21}$
(B) $2 \frac{4}{21}$
(C) $2 \frac{5}{21}$
(D) $2 \frac{6}{21}$

Part B

Problem 11

In the adjoining figure $\angle \mathrm{BAD}=\angle \mathrm{DAF}=\angle \mathrm{FAC}$. GE is parallel to $\mathrm{DF}$, and $\angle \mathrm{EGA}=90^{\circ}$. If $\angle \mathrm{ACE}=70^{\circ}$, the measure of $\angle \mathrm{FDE}$ is $\rule{1cm}{0.15mm}$.

Problem 12

$A B C$ is a triangle in which the angles are in the ratio $3: 4: 5$. PQR is a triangle in which the angles are in the ratio $5: 6: 7$. The difference between the least angle of $A B C$ and the least angle of PQR is $a^{\circ}$. Then $a=$ $\rule{1cm}{0.15mm}$.

Problem 13

Samrud had to multiply a number by 35 . By mistake he multiplied by 53 and got a result 720 more. The new product is $\rule{1cm}{0.15mm}$.

Problem 14

Vishva plays football every 4 th day. He played on a Tuesday. He plays football on a Tuesday again in is $\rule{1cm}{0.15mm}$ days.

Problem 15

In an elementary school $26 \%$ of the students are girls. If there are 240 less girls than boys, then the strength of the school is $\rule{1cm}{0.15mm}$.

Problem 16

There are three concentric circles as shown in the figure. The radii of them are $2 \mathrm{~cm}, 4 \mathrm{~cm}$ and 6 $\mathrm{cm}$. The ratio of the area of the shaded region to the area of the dotted region is $\frac{a}{b}$ where $a, b$ are integers and have no common factor other than 1. Then $a+b=$ $\rule{1cm}{0.15mm}$.

Problem 17

The value of $\left(1+\frac{1}{9}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{7}\right)\left(1+\frac{1}{6}\right)\left(1+\frac{1}{5}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{2}\right)$ is $\rule{1cm}{0.15mm}$.

Problem 18

When a two digit number divides 265 , the remainder is 5 . The number of such two digit numbers is $\rule{1cm}{0.15mm}$.

Problem 19

If $A \# B=\frac{A \times B}{A+B}$, the value of $\frac{12 \# 8}{8 \# 4}+\frac{10 \# 6}{6 \# 2}$ is $\rule{1cm}{0.15mm}$.

Problem 20

When water becomes ice, its volume increases by $10 \%$. When ice melts into water its volume decreases by $a \%$. Then $a=$ $\rule{1cm}{0.15mm}$.