NMTC GAUSS Primary Level 2025 Solved Question Paper PDF

Every term in the numerator and in the denominator is a multiple of 1111. Taking 1111 common, the numerator is $$1111 \times (9 + 7 + 5) = 1111 \times 21$$ and the denominator is $$1111 \times (8 + 6 + 4) = 1111 \times 18$$. Cancelling 1111 from both gives $$\frac{21}{18} = \frac{7}{6}$$.

The sum of three odd numbers is always odd, but 30 is even, so one of the three primes must be the only even prime 2. The other two primes must then add up to 28, and the only prime pairs with sum 28 are 5 and 23, and 11 and 17. So the sets are $$\{2, 5, 23\}$$ and $$\{2, 11, 17\}$$, which means there are 2 such sets.

The figure marks $$\angle EAD = 20^\circ$$, and since $$AD$$ bisects $$\angle EAB$$, we get $$\angle EAB = 2 \times 20^\circ = 40^\circ$$. The triangle $$ABC$$ is equilateral, so $$\angle BAC = 60^\circ$$ and hence $$\angle EAC = 40^\circ + 60^\circ = 100^\circ$$. As $$\ell_1$$ and $$\ell_2$$ are parallel with $$AC$$ as the transversal, $$x$$ and $$\angle EAC$$ are alternate angles, so $$x = 100^\circ$$.

Every piece of the square has an area of $$1\ \text{cm}^2$$ or $$2\ \text{cm}^2$$, so the area of the whole square must be a whole number. If the perimeter is $$P$$, the side is $$\frac{P}{4}$$ and the area is $$\frac{P^2}{16}$$. Now $$\frac{17^2}{16}$$, $$\frac{15^2}{16}$$ and $$\frac{14^2}{16}$$ are not whole numbers, while $$\frac{16^2}{16} = 16$$ is. So the side is 4 cm and the perimeter is 16 cm.

Using the given rule, $$13 \ast 6 = \frac{13 + 6}{13 - 6} = \frac{19}{7}$$ and $$5 \ast 2 = \frac{5 + 2}{5 - 2} = \frac{7}{3}$$. Dividing one by the other, $$\frac{19}{7} \div \frac{7}{3} = \frac{19}{7} \times \frac{3}{7} = \frac{57}{49}$$.

Every corner of the shaded figure sits on a dot of the grid, so the dot rule for area can be used, which says the area equals the number of dots strictly inside, plus half the number of dots on the boundary, minus 1. This figure has 6 dots inside it and 21 dots on its boundary, so the area is $$6 + \frac{21}{2} - 1 = \frac{31}{2}$$ square cm.

The three natural numbers are different and increasing, so the smallest they can possibly be is 1, 2 and 3, and these already add up to 6. Any other set of three different natural numbers has a sum bigger than 6. So $$n_1 = 1$$, $$n_2 = 2$$, $$n_3 = 3$$ and the value of $$n_3$$ is 3.

In the triangle $$ABC$$, the angle $$\angle ACB$$ is vertically opposite the marked $$70^\circ$$, so $$\angle BAC = 180^\circ - 30^\circ - 70^\circ = 80^\circ$$ and the bisector $$AP$$ splits it into two angles of $$40^\circ$$. Where $$AP$$ crosses the line $$BD$$, the angle formed with $$BD$$ is $$180^\circ - 30^\circ - 40^\circ = 110^\circ$$, that is $$70^\circ$$ on the other side. In the four sided figure $$CDEF$$ the angles at $$D$$ and at $$F$$ are right angles and the angle at $$C$$ is $$70^\circ$$, so $$\angle DEF = 360^\circ - 70^\circ - 90^\circ - 90^\circ = 110^\circ$$, and its bisector $$EQ$$ makes $$55^\circ$$ with $$ED$$ and therefore $$35^\circ$$ with $$BD$$. The two bisectors together with $$BD$$ form a triangle, so the angle between the bisectors is $$180^\circ - 70^\circ - 35^\circ = 75^\circ$$, and $$x$$ is vertically opposite that angle.

The numbers 70 to 79 have 7 in the tens place, and that is 10 numbers. The two-digit numbers ending in 7 are 17, 27, 37, 47, 57, 67, 87 and 97, which is 8 more, because 77 has already been counted. So the total is $$10 + 8 = 18$$.

The gap between the two marked fractions is $$\frac{1}{3} - \frac{1}{5} = \frac{2}{15}$$, and the number line cuts this gap into 16 equal parts, so each small part is $$\frac{2}{15} \div 16 = \frac{1}{120}$$. Since $$\frac{1}{4} - \frac{1}{5} = \frac{1}{20} = \frac{6}{120}$$, the fraction $$\frac{1}{4}$$ lies 6 small parts to the right of $$\frac{1}{5}$$, which is the mark $$q$$.

Take the whole book as 1. After the first day $$\frac{2}{3}$$ is left. On the second day he reads half of that, which is $$\frac{1}{3}$$ of the book, so $$\frac{1}{3}$$ of the book is still left. On the third day he reads $$\frac{1}{4}$$ of $$\frac{2}{3}$$, which is $$\frac{1}{6}$$ of the book. The unread part is $$\frac{1}{3} - \frac{1}{6} = \frac{1}{6}$$ of the book, and this equals 23 pages, so the book has $$6 \times 23 = 138$$ pages.

Since $$100 = 2^2 \times 5^2$$, the only way of writing 100 as a product of four different natural numbers is $$1 \times 2 \times 5 \times 10$$. Their sum is $$1 + 2 + 5 + 10 = 18$$.

Walking 100 m East and later 70 m West leaves him $$100 - 70 = 30$$ m East of A. Walking 30 m North and later 10 m North leaves him $$30 + 10 = 40$$ m North of A. So AB is the longest side of a right angled triangle with the other two sides 30 m and 40 m, giving $$AB = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50$$ m.

The figure marks $$\angle ADC = 92^\circ$$, so $$\angle DAB = 92^\circ + 8^\circ = 100^\circ$$ and $$\angle BCD = 92^\circ - 8^\circ = 84^\circ$$. The four angles of $$ABCD$$ add up to $$360^\circ$$, so $$\angle ABC = 360^\circ - 92^\circ - 100^\circ - 84^\circ = 84^\circ$$. As $$E$$ lies on $$AB$$ produced and $$F$$ lies on $$CB$$ produced, $$\angle FBE$$ is vertically opposite $$\angle ABC$$ and so equals $$84^\circ$$. Then $$\angle FEB = \frac{84^\circ}{2} = 42^\circ$$ and $$\angle BFE = 180^\circ - 84^\circ - 42^\circ = 54^\circ$$.

Simplify from the bottom upwards. First $$1 + \frac{1}{4} = \frac{5}{4}$$, so the next step is $$3 + \frac{4}{5} = \frac{19}{5}$$, and then $$2 + \frac{5}{19} = \frac{43}{19}$$. The given fraction is therefore $$\frac{19}{43}$$, and the fraction to be added is $$1 - \frac{19}{43} = \frac{24}{43}$$.

The shares of A, B and C are in the ratio $$100 \colon 65 \colon 40$$, which is $$100 + 65 + 40 = 205$$ parts altogether. C has 40 parts and that is Rs. 4000, so one part is Rs. 100. The total amount is $$205 \times 100 = 20500$$ rupees.

The five angles of a pentagon add up to $$540^\circ$$ and the ratio has $$8 + 9 + 12 + 15 + 10 = 54$$ parts, so one part is $$10^\circ$$. This makes $$\angle B = 90^\circ$$ and $$\angle C = 120^\circ$$. The exterior angle at B is $$180^\circ - 90^\circ = 90^\circ$$, so its bisector makes $$45^\circ$$ with BC, and the internal bisector of $$\angle C$$ makes $$60^\circ$$ with CB. In the triangle BPC, $$\angle BPC = 180^\circ - 45^\circ - 60^\circ = 75^\circ$$.

The number reduced by 5 must be a common multiple of 36, 48, 21 and 28, so the smallest such value is their LCM. Since $$36 = 2^2 \times 3^2$$, $$48 = 2^4 \times 3$$, $$21 = 3 \times 7$$ and $$28 = 2^2 \times 7$$, the LCM is $$2^4 \times 3^2 \times 7 = 1008$$. Adding back the 5 gives $$1008 + 5 = 1013$$.

First $$10\frac{5}{6} = \frac{65}{6}$$, so dividing by 91 gives $$\frac{65}{6} \times \frac{1}{91} = \frac{65}{546}$$. Since $$65 = 5 \times 13$$ and $$546 = 2 \times 3 \times 7 \times 13$$, cancelling 13 leaves $$\frac{5}{42}$$ in lowest terms. So $$a = 5$$, $$b = 42$$ and $$b - a = 37$$.

Trying the smallest primes, $$p = 2$$ fails because $$2 + 6 = 8$$ is not prime, and $$p = 3$$ fails because $$3 + 6 = 9$$ is not prime. For $$p = 5$$ the four numbers are 11, 13, 17 and 19, all of which are prime, so $$p = 5$$. Then $$p^2 = 25$$ and $$25 = 4 \times 6 + 1$$, so the remainder is 1.

Let the total number of balls initially be $$x$$.

Since $$90\%$$ of the balls are black,

$$\text{Number of black balls}=90\% \text{ of }x=\frac{90}{100}x=\frac{9x}{10}.$$

Therefore, the number of white balls initially is

$$x-\frac{9x}{10}=\frac{x}{10}.$$

When 9 white balls are added,

$$\text{Number of black balls}=\frac{9x}{10}$$

and

$$\text{Number of white balls}=\frac{x}{10}+9.$$

According to the question, the ratio of black balls to white balls is $$4:3$$.

Therefore,

$$\frac{\frac{9x}{10}}{\frac{x}{10}+9}=\frac{4}{3}.$$

Cross-multiplying,

$$3\left(\frac{9x}{10}\right)= 4\left(\frac{x}{10}+9\right).$$

$$\frac{27x}{10} = \frac{4x}{10}+36.$$

Multiplying throughout by $$10$$,

$$27x=4x+360.$$

$$23x=360.$$

$$x=\frac{360}{23}.$$

But the number of balls must be a whole number. Hence, the given data are inconsistent.

Therefore, $$\textbf{there is no possible whole-number solution for the number of white balls initially}.$$

$${\text{No valid whole-number solution}}$$

Three straight lines pass through the middle point of the figure and cut out three triangles, one carrying the angles $$a$$ and $$b$$, one carrying $$e$$ and $$f$$, and one carrying $$c$$ and $$d$$. The three angles of these triangles at the middle point are vertically opposite the three remaining angles around that point, so the three of them add up to half of $$360^\circ$$, which is $$180^\circ$$. All nine angles of the three triangles add up to $$3 \times 180^\circ = 540^\circ$$, so the six marked angles add up to $$540^\circ - 180^\circ = 360^\circ$$.

Adding the three given totals counts every fruit exactly twice, so twice the number of fruits is $$88 + 80 + 64 = 232$$, and the basket holds 116 fruits in all. The bananas and oranges together are 64, so the number of apples is $$116 - 64 = 52$$.

Since $$AB = AC$$ and the figure marks $$\angle ABC = 50^\circ$$, we get $$\angle ACB = 50^\circ$$ and $$\angle BAC = 180^\circ - 50^\circ - 50^\circ = 80^\circ$$. As $$FD$$ is parallel to $$AC$$ with $$AB$$ as the transversal, the corresponding angles give $$\angle BDF = \angle BAC = 80^\circ$$, and since $$E$$ lies on $$DB$$, this is the angle $$\angle EDF$$ of the triangle $$EDF$$. In that triangle $$EF = DE$$, so the angles opposite these equal sides are equal, which gives $$x = \angle DFE = \angle EDF = 80^\circ$$.

Half the perimeter is $$\frac{40}{2} = 20$$ cm, so the length and the breadth are two primes adding up to 20. The possible prime pairs are 3 and 17, giving an area of $$3 \times 17 = 51$$, and 7 and 13, giving an area of $$7 \times 13 = 91$$. The greater of the two is $$91\ \text{cm}^2$$.